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Precalculus Quiz

Precalculus Quiz: Constructing Tangents To Circles

Practice Constructing Tangents To Circles in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

From point $Q(0, 8)$ , tangent lines are constructed to the circle $(x-3)^2 + (y-2)^2 = 13$ . What is the angle between the two tangent lines?

What this quiz covers

This quiz focuses on Constructing Tangents To Circles, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

From point $Q(0, 8)$ , tangent lines are constructed to the circle $(x-3)^2 + (y-2)^2 = 13$ . What is the angle between the two tangent lines?

$30°$
$45°$
$60°$ (correct answer)
$90°$

Explanation: The circle has center $(3,2)$ and radius $\sqrt{13}$ . Distance from $Q$ to center is $\sqrt{9+36} = 3\sqrt{5}$ . In the right triangle formed by the center, external point, and tangent point, $\sin(\alpha) = \frac{r}{d} = \frac{\sqrt{13}}{3\sqrt{5}} = \sqrt{\frac{13}{45}}$ . Since $\frac{13}{45} = \frac{13}{45} \approx 0.289$ , we get $\sin(\alpha) \approx 0.538$ , so $\alpha \approx 30°$ . The angle between the tangents is $2\alpha = 60°$ .

Question 2

A circle with equation $x^2 + y^2 - 6x + 4y - 12 = 0$ is given. From which of the following points can exactly one tangent line be drawn to this circle?

$(8, 2)$ only
$(3, -2)$ only
Any point at distance $5$ from $(3, -2)$ (correct answer)
Any point at distance $\sqrt{37}$ from $(3, -2)$

Explanation: First, rewrite the circle in standard form: $(x-3)^2 + (y+2)^2 = 25$ , so center is $(3,-2)$ and radius is $5$ . Exactly one tangent can be drawn from a point if and only if the point lies on the circle itself. Points on the circle are at distance $5$ from the center $(3,-2)$ . Choice A is a specific external point, choice B is the center, and choice D represents points outside the circle.

Question 3

A tangent to the circle $(x-h)^2 + (y-k)^2 = r^2$ is constructed from external point $(a,b)$ . The tangent line passes through point $(2a-h, 2b-k)$ . Which relationship must be true?

The point $(2a-h, 2b-k)$ is the reflection of $(a,b)$ across the center
The point $(a,b)$ is equidistant from $(h,k)$ and $(2a-h, 2b-k)$ (correct answer)
The tangent line is the perpendicular bisector of the segment from $(h,k)$ to $(a,b)$
The distance from $(h,k)$ to the tangent line equals $\sqrt{(a-h)^2 + (b-k)^2}$

Explanation: The point $(2a-h, 2b-k)$ can be written as $(a,b) + ((a,b) - (h,k))$ , which means $(a,b)$ is the midpoint of the segment from $(h,k)$ to $(2a-h, 2b-k)$ . Therefore, $(a,b)$ is equidistant from these two points. Choice A incorrectly describes the reflection relationship, choice C incorrectly identifies the tangent line's role, and choice D incorrectly states that the distance to the tangent equals the distance to the external point.

Question 4

A circle $O$ has center $O$ and radius $5$ . An external point $P$ is such that $OP=13$ . From $P$ , a tangent segment $PT$ is drawn to the circle, touching the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$8$
$12$ (correct answer)
$18$
$\sqrt{194}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 5 and the external point P is at distance d = 13 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 5² = 13², so t² + 25 = 169, thus t² = 144 and t = √144 = 12. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice A incorrectly reverses the Pythagorean relationship, treating the tangent length as the hypotenuse when actually the distance OP is the hypotenuse. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 5

Circle $O$ has center $O$ and radius $8$ . Point $P$ is outside the circle with $OP=17$ . Tangent segments $PT$ and $PU$ are drawn from $P$ to the circle, touching at $T$ and $U$ .

Which of the following is true about the two tangent segments from $P$ ?

$PT < PU$
$PT = PU$ (correct answer)
$PT > PU$
One of $PT$ and $PU$ must be a diameter of the circle

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d, same radius r), both have length t = √(d² - r²), making them equal; additionally, the configuration is symmetric about line OP. Choice B is correct because it correctly identifies equal lengths. Choice A incorrectly claims the two tangent segments have different lengths, but by symmetry and the Pythagorean theorem with the same parameters, they must be equal. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 6

A circle $O$ has center $O$ and radius $6$ . An external point $P$ satisfies $OP=10$ . A tangent from $P$ touches the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$16$
$8$ (correct answer)
$4$
$\sqrt{136}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length ( $t$ ), radius ( $r$ ), and distance from center to external point ( $d$ ) forms a right triangle where $t^2 + r^2 = d^2$ by the Pythagorean theorem. Given that the circle has radius $r = 6$ and the external point $P$ is at distance $d = 10$ from center $O$ , we form a right triangle with the radius as one leg, the tangent segment as the other leg, and $OP$ as the hypotenuse, giving us $t^2 + 6^2 = 10^2$ , so $t^2 + 36 = 100$ , thus $t^2 = 64$ and $t = \sqrt{64} = 8$ . Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice A incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: $d \neq r + t$ , rather $d^2 = r^2 + t^2$ . Key to tangent problems: remember that tangent $\perp$ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius $r$ and distance $d$ to external point, then use $t = \sqrt{d^2 - r^2}$ ; if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 7

Circle $O$ has radius $3$ , and an external point $P$ satisfies $OP=5$ . Tangent segments from $P$ touch the circle at points of tangency. Using the given circle and point, how many tangent lines can be drawn from point $P$ to circle $O$ ?

$0$
$1$
$2$ (correct answer)
Infinitely many

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. From external point P, we can construct exactly two tangent lines because as we rotate a line through P, there are exactly two positions where the line is perpendicular to a radius, and those positions define the two tangent lines. Choice C is correct because it correctly counts tangents. Choice B incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 8

Circle $O$ has center $O$ and radius $4$ . Point $P$ is outside the circle, and a tangent segment $PT$ touches the circle at $T$ . To construct a tangent from $P$ to circle $O$ , what must be true at the point of tangency $T$ ?

The segment $OT$ is perpendicular to the tangent line $PT$ . (correct answer)
The segment $OT$ is parallel to the tangent line $PT$ .
The point $T$ must lie on line $OP$ .
The segment $PT$ must pass through the center $O$ .

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. At the point of tangency, the radius and tangent line meet at a right angle (90°), which is the defining property of a tangent line. The tangent line must be perpendicular to the radius OT at point T because this is the definition of tangency—any line through T that is not perpendicular to OT would either miss the circle entirely or intersect it at a second point. Choice A is correct because it correctly states the perpendicularity property that defines tangency. Choice B incorrectly claims the radius and tangent are parallel, but the defining property of tangency is perpendicularity. When constructing tangents, the perpendicularity condition is essential: the tangent line must be perpendicular to the radius at the point where it touches the circle.

Question 9

A circle is tangent to both coordinate axes in the first quadrant. If a tangent line from point $(8, 0)$ to this circle has slope $m = -\frac{3}{4}$ , what is the radius of the circle?

$2$
$3$ (correct answer)
$4$
$5$

Explanation: Since the circle is tangent to both axes in the first quadrant, its center is at $(r,r)$ where $r$ is the radius. The tangent line from $(8,0)$ with slope $-\frac{3}{4}$ has equation $y = -\frac{3}{4}(x-8) = -\frac{3}{4}x + 6$ . The distance from center $(r,r)$ to this line must equal the radius: $\frac{|\frac{3}{4}r + r - 6|}{\sqrt{\frac{9}{16} + 1}} = r$ . This gives $\frac{|\frac{7}{4}r - 6|}{\frac{5}{4}} = r$ , so $|\frac{7}{4}r - 6| = \frac{5}{4}r$ . Since $r < 6$ , we have $6 - \frac{7}{4}r = \frac{5}{4}r$ , giving $6 = 3r$ , so $r = 3$ .

Question 10

Circle $O$ has center $O$ and radius $4$ . Point $P$ is an external point with $OP=5$ . Two tangent segments from $P$ touch the circle at $T_1$ and $T_2$ . For the circle described, which of the following is true about the two tangent segments $PT_1$ and $PT_2$ ?

$PT_1 = PT_2$ (correct answer)
$PT_1 > PT_2$
$PT_1 < PT_2$
There is only one tangent segment from $P$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d = 5, same radius r = 4), both have length t = √(d² - r²) = √(25 - 16) = √9 = 3, making them equal; additionally, the configuration is symmetric about line OP. Choice A is correct because it correctly identifies equal lengths. Choice D incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration.

Question 11

A circle $O$ has center $O(0,0)$ and radius $5$ . Point $P$ is located at $(13,0)$ , and a tangent segment $PT$ is drawn from $P$ to the circle, touching the circle at point $T$ . For the circle described, what is the length of the tangent segment $PT$ ?

$18$
$12$ (correct answer)
$8$
$ext{13}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. A tangent line to a circle is a line that touches the circle at exactly one point, called the point of tangency, and is perpendicular to the radius at that point. Given that the circle has radius r = 5 and the external point P is at distance d = 13 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 5² = 13², so t² + 25 = 169, thus t² = 144 and t = √144 = 12. Choice B is correct because it applies the Pythagorean theorem correctly with the given values. Choice D incorrectly uses 13 as the tangent length, but 13 is actually the distance from O to P, not the tangent length. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

Question 12

Circle $O$ has center $O$ and radius $4$ . An external point $P$ satisfies $OP=10$ , and tangents from $P$ touch the circle at $T$ and $U$ . For the circle described, which of the following is true about the two tangent segments $PT$ and $PU$ ?

$PT > PU$
$PT < PU$
$PT = PU$ (correct answer)
Their lengths cannot be compared without coordinates.

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d, same radius r), both have length t = √(d² - r²), making them equal; additionally, the configuration is symmetric about line OP. Choice C is correct because it correctly identifies equal lengths. Choice D incorrectly claims their lengths cannot be compared without coordinates, but by symmetry and the Pythagorean theorem with the same parameters, they must be equal. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

Question 13

Circle $O$ has center $O$ and radius $6$ . Point $P$ is an external point with $OP=10$ . Two tangent lines can be drawn from $P$ to the circle.

For the circle described, how many tangent lines can be drawn from point $P$ to circle $O$ ?

$0$
$1$
$2$ (correct answer)
Infinitely many

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. From external point P, we can construct exactly two tangent lines because as we rotate a line through P, there are exactly two positions where the line is perpendicular to a radius, and those positions define the two tangent lines. Choice C is correct because it correctly counts tangents. Choice B incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 14

A circle $O$ has center $O$ and radius $7$ . An external point $P$ satisfies $OP=25$ . A tangent segment $PT$ touches the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$24$ (correct answer)
$18$
$32$
$\sqrt{674}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. Choice A is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 15

Circle $O$ has center $O$ and radius $7$ . Point $P$ is an external point with $OP=25$ , and a tangent segment $PT$ touches the circle at $T$ . For the circle described, what is the length of the tangent segment $PT$ ?

$18$
$24$ (correct answer)
$26$
$32$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly uses 26, which would be the hypotenuse if the tangent were 24 and radius were 10, mixing up the problem parameters. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 16

Circle $O$ has radius $4$ , and an external point $P$ satisfies $OP=5$ . Tangent segments from $P$ touch the circle at $T$ and $T'$ . Using the given circle and point, what is the length of the tangent segment $PT$ ?

$1$
$3$ (correct answer)
$9$
$\sqrt{41}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 4 and the external point P is at distance d = 5 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 4² = 5², so t² + 16 = 25, thus t² = 9 and t = √9 = 3. This is a 3-4-5 Pythagorean triple, so we immediately recognize the tangent length is 3. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and distance instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 17

A circle $O$ has center $O$ and radius $4$ . Point $P$ lies outside the circle with $OP=5$ . Tangent segments from $P$ touch the circle at points $T$ and $U$ .

Using the given circle and point, what is the length of the tangent segment from $P$ to the circle (i.e., $PT$ )?

$3$ (correct answer)
$\sqrt{41}$
$9$
$1$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 4 and the external point P is at distance d = 5 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 4² = 5², so t² + 16 = 25, thus t² = 9 and t = √9 = 3. Choice A is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 18

A circle $O$ has center $O$ and radius $8$ . An external point $P$ satisfies $OP=17$ . A tangent from $P$ touches the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$9$
$15$ (correct answer)
$25$
$\sqrt{353}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 8 and the external point P is at distance d = 17 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 8² = 17², so t² + 64 = 289, thus t² = 225 and t = √225 = 15. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly reverses the Pythagorean relationship, treating the tangent length as the hypotenuse when actually the distance OP is the hypotenuse. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 19

Circle $O$ has center $O$ and radius $7$ . Point $P$ is an external point with $OP=25$ . Tangent segment $PT$ touches the circle at $T$ . For the circle described, what is the length of the tangent segment $PT$ ?

$18$
$24$ (correct answer)
$26$
$\sqrt{674}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. This is a 7-24-25 Pythagorean triple, so we immediately recognize the tangent length is 24. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and distance instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 20

Circle $O$ has center $O$ and radius $5$ . Point $P$ is an external point, and a tangent from $P$ touches the circle at $T$ .

To construct a tangent from $P$ to circle $O$ , what must be true at the point of tangency $T$ ?

The segment $OT$ is perpendicular to the line $PT$ (correct answer)
The segment $OT$ is parallel to the line $PT$
The segment $PT$ must pass through the center $O$
The point $T$ must be the midpoint of segment $OP$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. A tangent line to a circle is a line that touches the circle at exactly one point, called the point of tangency, and is perpendicular to the radius at that point. The tangent line must be perpendicular to the radius OT at point T because this is the definition of tangency—any line through T that is not perpendicular to OT would either miss the circle entirely or intersect it at a second point. Choice A is correct because it correctly states the perpendicularity property. Choice B claims the tangent and radius are parallel or at an angle other than 90°, but the defining property of tangency is perpendicularity. When constructing tangents, the perpendicularity condition is essential: the tangent line must be perpendicular to the radius at the point where it touches the circle. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

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Precalculus Quiz

Precalculus Quiz: Constructing Tangents To Circles

Practice Constructing Tangents To Circles in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

From point $Q(0, 8)$ , tangent lines are constructed to the circle $(x-3)^2 + (y-2)^2 = 13$ . What is the angle between the two tangent lines?

What this quiz covers

This quiz focuses on Constructing Tangents To Circles, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

From point $Q(0, 8)$ , tangent lines are constructed to the circle $(x-3)^2 + (y-2)^2 = 13$ . What is the angle between the two tangent lines?

$30°$
$45°$
$60°$ (correct answer)
$90°$

Question 2

A circle with equation $x^2 + y^2 - 6x + 4y - 12 = 0$ is given. From which of the following points can exactly one tangent line be drawn to this circle?

$(8, 2)$ only
$(3, -2)$ only
Any point at distance $5$ from $(3, -2)$ (correct answer)
Any point at distance $\sqrt{37}$ from $(3, -2)$

Question 3

A tangent to the circle $(x-h)^2 + (y-k)^2 = r^2$ is constructed from external point $(a,b)$ . The tangent line passes through point $(2a-h, 2b-k)$ . Which relationship must be true?

The point $(2a-h, 2b-k)$ is the reflection of $(a,b)$ across the center
The point $(a,b)$ is equidistant from $(h,k)$ and $(2a-h, 2b-k)$ (correct answer)
The tangent line is the perpendicular bisector of the segment from $(h,k)$ to $(a,b)$
The distance from $(h,k)$ to the tangent line equals $\sqrt{(a-h)^2 + (b-k)^2}$

Question 4

A circle $O$ has center $O$ and radius $5$ . An external point $P$ is such that $OP=13$ . From $P$ , a tangent segment $PT$ is drawn to the circle, touching the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$8$
$12$ (correct answer)
$18$
$\sqrt{194}$

Question 5

Circle $O$ has center $O$ and radius $8$ . Point $P$ is outside the circle with $OP=17$ . Tangent segments $PT$ and $PU$ are drawn from $P$ to the circle, touching at $T$ and $U$ .

Which of the following is true about the two tangent segments from $P$ ?

$PT < PU$
$PT = PU$ (correct answer)
$PT > PU$
One of $PT$ and $PU$ must be a diameter of the circle

Question 6

A circle $O$ has center $O$ and radius $6$ . An external point $P$ satisfies $OP=10$ . A tangent from $P$ touches the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$16$
$8$ (correct answer)
$4$
$\sqrt{136}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length ( $t$ ), radius ( $r$ ), and distance from center to external point ( $d$ ) forms a right triangle where $t^2 + r^2 = d^2$ by the Pythagorean theorem. Given that the circle has radius $r = 6$ and the external point $P$ is at distance $d = 10$ from center $O$ , we form a right triangle with the radius as one leg, the tangent segment as the other leg, and $OP$ as the hypotenuse, giving us $t^2 + 6^2 = 10^2$ , so $t^2 + 36 = 100$ , thus $t^2 = 64$ and $t = \sqrt{64} = 8$ . Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice A incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: $d \neq r + t$ , rather $d^2 = r^2 + t^2$ . Key to tangent problems: remember that tangent $\perp$ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius $r$ and distance $d$ to external point, then use $t = \sqrt{d^2 - r^2}$ ; if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 7

$0$
$1$
$2$ (correct answer)
Infinitely many

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. From external point P, we can construct exactly two tangent lines because as we rotate a line through P, there are exactly two positions where the line is perpendicular to a radius, and those positions define the two tangent lines. Choice C is correct because it correctly counts tangents. Choice B incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 8

The segment $OT$ is perpendicular to the tangent line $PT$ . (correct answer)
The segment $OT$ is parallel to the tangent line $PT$ .
The point $T$ must lie on line $OP$ .
The segment $PT$ must pass through the center $O$ .

Question 9

A circle is tangent to both coordinate axes in the first quadrant. If a tangent line from point $(8, 0)$ to this circle has slope $m = -\frac{3}{4}$ , what is the radius of the circle?

$2$
$3$ (correct answer)
$4$
$5$

Question 10

$PT_1 = PT_2$ (correct answer)
$PT_1 > PT_2$
$PT_1 < PT_2$
There is only one tangent segment from $P$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d = 5, same radius r = 4), both have length t = √(d² - r²) = √(25 - 16) = √9 = 3, making them equal; additionally, the configuration is symmetric about line OP. Choice A is correct because it correctly identifies equal lengths. Choice D incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration.

Question 11

$18$
$12$ (correct answer)
$8$
$ext{13}$

Question 12

$PT > PU$
$PT < PU$
$PT = PU$ (correct answer)
Their lengths cannot be compared without coordinates.

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d, same radius r), both have length t = √(d² - r²), making them equal; additionally, the configuration is symmetric about line OP. Choice C is correct because it correctly identifies equal lengths. Choice D incorrectly claims their lengths cannot be compared without coordinates, but by symmetry and the Pythagorean theorem with the same parameters, they must be equal. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

Question 13

Circle $O$ has center $O$ and radius $6$ . Point $P$ is an external point with $OP=10$ . Two tangent lines can be drawn from $P$ to the circle.

For the circle described, how many tangent lines can be drawn from point $P$ to circle $O$ ?

$0$
$1$
$2$ (correct answer)
Infinitely many

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. From external point P, we can construct exactly two tangent lines because as we rotate a line through P, there are exactly two positions where the line is perpendicular to a radius, and those positions define the two tangent lines. Choice C is correct because it correctly counts tangents. Choice B incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 14

A circle $O$ has center $O$ and radius $7$ . An external point $P$ satisfies $OP=25$ . A tangent segment $PT$ touches the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$24$ (correct answer)
$18$
$32$
$\sqrt{674}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. Choice A is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 15

$18$
$24$ (correct answer)
$26$
$32$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly uses 26, which would be the hypotenuse if the tangent were 24 and radius were 10, mixing up the problem parameters. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 16

$1$
$3$ (correct answer)
$9$
$\sqrt{41}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 4 and the external point P is at distance d = 5 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 4² = 5², so t² + 16 = 25, thus t² = 9 and t = √9 = 3. This is a 3-4-5 Pythagorean triple, so we immediately recognize the tangent length is 3. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and distance instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 17

A circle $O$ has center $O$ and radius $4$ . Point $P$ lies outside the circle with $OP=5$ . Tangent segments from $P$ touch the circle at points $T$ and $U$ .

Using the given circle and point, what is the length of the tangent segment from $P$ to the circle (i.e., $PT$ )?

$3$ (correct answer)
$\sqrt{41}$
$9$
$1$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 4 and the external point P is at distance d = 5 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 4² = 5², so t² + 16 = 25, thus t² = 9 and t = √9 = 3. Choice A is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 18

A circle $O$ has center $O$ and radius $8$ . An external point $P$ satisfies $OP=17$ . A tangent from $P$ touches the circle at $T$ .

For the circle described, what is the length of the tangent segment $PT$ ?

$9$
$15$ (correct answer)
$25$
$\sqrt{353}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 8 and the external point P is at distance d = 17 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 8² = 17², so t² + 64 = 289, thus t² = 225 and t = √225 = 15. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly reverses the Pythagorean relationship, treating the tangent length as the hypotenuse when actually the distance OP is the hypotenuse. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 19

$18$
$24$ (correct answer)
$26$
$\sqrt{674}$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. This is a 7-24-25 Pythagorean triple, so we immediately recognize the tangent length is 24. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and distance instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 20

Circle $O$ has center $O$ and radius $5$ . Point $P$ is an external point, and a tangent from $P$ touches the circle at $T$ .

To construct a tangent from $P$ to circle $O$ , what must be true at the point of tangency $T$ ?

The segment $OT$ is perpendicular to the line $PT$ (correct answer)
The segment $OT$ is parallel to the line $PT$
The segment $PT$ must pass through the center $O$
The point $T$ must be the midpoint of segment $OP$

Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. A tangent line to a circle is a line that touches the circle at exactly one point, called the point of tangency, and is perpendicular to the radius at that point. The tangent line must be perpendicular to the radius OT at point T because this is the definition of tangency—any line through T that is not perpendicular to OT would either miss the circle entirely or intersect it at a second point. Choice A is correct because it correctly states the perpendicularity property. Choice B claims the tangent and radius are parallel or at an angle other than 90°, but the defining property of tangency is perpendicularity. When constructing tangents, the perpendicularity condition is essential: the tangent line must be perpendicular to the radius at the point where it touches the circle. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.