Circles - PSAT Math

A line is tangent to a circle at point $P$ . The circle has center $O$ , and $OP=13$ . If the distance from $O$ to the tangent line is $13$ , what is the measure of the angle between $\overline{OP}$ and the tangent line at $P$ ?
Answer: $90^\circ$ This question involves a fundamental property of tangent lines to circles. A tangent line to a circle is always perpendicular to the radius drawn to the point of tangency. Since OP is the radius to point P where the line is tangent, the angle between OP and the tangent line must be 90°. The given information that the distance from O to the tangent line equals OP = 13 confirms this is the perpendicular distance. Remember this key property for any tangent line problem.
In circle $O$ , chord $AB$ has length $16$ . The perpendicular distance from the center $O$ to chord $AB$ is $6$ . What is the radius of the circle?
Answer: $10$ This question asks for the radius of circle O given chord AB of length 16 and a perpendicular distance of 6 from center O to the chord. The property states that the perpendicular from the center to a chord bisects the chord, so half the chord is 8, forming a right triangle with legs 6 and 8, and hypotenuse as the radius r. Using the Pythagorean theorem, $r^2$ = $6^2$ + $8^2$ = 36 + 64 = 100, so r = √100 = 10. Draw a diagram to visualize the right triangle with the radius as hypotenuse. A frequent error is forgetting to halve the chord, leading to incorrect Pythagorean application. For strategy, emphasize using radius in the hypotenuse and verify by checking if the distance is less than the radius.
In the figure shown, a circle is inscribed in a right triangle with legs of lengths $6$ and $8$ . What is the radius of the inscribed circle?
Answer: $2$ For a right triangle with legs $a$ , $b$ and hypotenuse $c$ , the inradius is $r = \frac{a+b-c}{2}$ . Here $c = \sqrt{36+64} = 10$ , so $r = \frac{6+8-10}{2} = 2$ . Alternative: $r = \text{Area}/s = 24/12 = 2$ . Choice A comes from dividing by 4 instead of 2. Choice C is the altitude to the hypotenuse ( $6 \cdot 8/10$ ). Choice D is half a leg.
Refer to the figure. In circle $R$ , arcs $AB$ , $BC$ , and $CA$ are shown. If $m\widehat{AB}=110^{\circ}$ and $m\widehat{BC}=95^{\circ}$ , what is $m\widehat{CA}$ ?
Answer: $155^{\circ}$ The three arcs together complete the full circle, so their measures must sum to $360^{\circ}$ . Therefore: $m\widehat{AB} + m\widehat{BC} + m\widehat{CA} = 360^{\circ}$ . Substituting the given values: $110^{\circ} + 95^{\circ} + m\widehat{CA} = 360^{\circ}$ . Solving: $m\widehat{CA} = 360^{\circ} - 205^{\circ} = 155^{\circ}$ .
A tangent line touches a circle at point $T$ . The circle has center $O$ . If $OT=13$ and a point $P$ lies on the tangent line such that $PT=5$ , what is the length of $OP$ ? (Use the relationship between a radius and a tangent.)
Answer: $\sqrt{194}$ This question asks for the length of OP, where a tangent touches a circle at T, O is the center with OT = 13, and P is on the tangent with PT = 5. The key property is that a radius to the point of tangency is perpendicular to the tangent line, forming right triangle OTP at T. In this right triangle, OT is one leg (13, the radius), PT is the other leg (5), and OP is the hypotenuse. Apply the Pythagorean theorem: OP = sqrt( $OT^2$ + $PT^2$ ) = sqrt( $13^2$ + $5^2$ ) = sqrt(169 + 25) = sqrt(194). A common error is confusing radius with diameter; using diameter 26 would incorrectly give sqrt( $26^2$ + $5^2$ ). Another mistake is not recognizing the right angle at T. As a test-taking tip, recall that tangents are perpendicular to radii to quickly set up the right triangle.
In the coordinate plane, a circle has equation $(x-3)^2+(y+4)^2=49$ . What is the center of the circle?
Answer: $(3,-4)$ This question asks for the center of a circle given its equation in standard form. The standard form is (x-h)² + (y-k)² = r², where (h,k) is the center. From (x-3)² + (y+4)² = 49, we can identify h = 3 and k = -4, so the center is (3,-4). A common error is misreading the signs: since we have (y+4)², this means (y-(-4))², so k = -4, not 4. Remember that the standard form uses subtraction, so plus signs indicate negative coordinates.
Refer to the figure. In circle $K$ with radius $9\text{ cm}$ , central angle $\angle MKL$ measures $40^{\circ}$ . What is the area, in square centimeters, of sector $MKL$ ?
Answer: $9\pi$ Sector area: $\frac{\theta}{360^{\circ}}(\pi r^{2})=\frac{40}{360}(\pi\cdot9^{2})=\frac{1}{9}(81\pi)=9\pi$ square cm (A). B would result from using 120° instead of 40°, C from using a 45° angle, and D gives the full circle area.
Refer to the figure. In circle $O$ , arc $AB = 80°$ and arc $CD = 30°$ , where chords $AC$ and $BD$ extended meet at external point $P$ . What is the measure of $\angle APB$ ?
Answer: $25°$ The angle formed by two secants meeting outside the circle equals half the positive difference of the intercepted arcs: $\angle APB = \frac{1}{2}(80° - 30°) = 25°$ . Choice A comes from dividing incorrectly. Choice C adds and halves ( $110/2$ ). Choice D is the sum of the arcs.
A circle has radius $12\text{ cm}$ . Point $A$ and point $B$ on the circle form a central angle of $150^\circ$ at the center. What is the length of arc $\overset{\frown}{AB}$ , in centimeters?
Answer: $10\pi\text{ cm}$ This question asks for the length of arc AB in a circle with radius 12 cm and a central angle of 150° between points A and B. The formula for arc length is (θ/360) × 2πr, where θ is the central angle in degrees and r is the radius. Substituting the values, arc length = (150/360) × 2π × 12 = (5/12) × 24π = 10π cm, using the correct radius of 12 cm. Simplify the fraction 150/360 to 5/12 before multiplying to avoid calculation errors. A key error to avoid is using the diameter instead of the radius in the formula, which would double the result incorrectly. For test-taking, remember to emphasize the radius in circumference calculations and check if the angle fraction is reduced properly.
A circle has center $(1,-5)$ and passes through $(1,2)$ . What is the area of the circle, in square units?
Answer: $49\pi$ This question asks for the area of a circle with center (1, -5) passing through (1, 2). First, find the radius as the distance between these points: √[ $(1-1)^2$ + $(2 - (-5))^2$ ] = √[0 + $7^2$ ] = 7. The area formula is π $r^2$ = π × 49 = 49π square units, using the correct radius. Note the vertical distance since x-coordinates match. A common error is squaring the radius incorrectly or using diameter. In exams, verify by substituting back to ensure the point lies on the circle.
A circular track has radius $50$ meters. A runner completes $\tfrac{3}{5}$ of a lap. What distance did the runner travel, in meters?
Answer: $60\pi$ This question asks for the arc length traveled by a runner on a circular track. The full circumference of the track is C = 2πr = 2π(50) = 100π meters. Since the runner completes 3/5 of a lap, the distance traveled is (3/5) × 100π = 60π meters. A common error is confusing the fraction of the lap with some other calculation or forgetting to multiply by the full circumference. When dealing with partial laps or arcs, always find the full circumference first, then multiply by the fraction.
In the figure shown, circle $O$ has radius $10$ . Chord $AB$ has length $16$ , and point $M$ is the midpoint of $AB$ . Point $P$ lies on the minor arc $AB$ such that $O$ , $M$ , and $P$ are collinear. What is the length of $MP$ ?
Answer: $4$ Since $M$ is the midpoint of chord $AB$ , $OM \perp AB$ and $AM = 8$ . By the Pythagorean theorem, $OM = \sqrt{10^2 - 8^2} = 6$ . Point $P$ on the minor arc lies at distance $10$ from $O$ on the line through $O$ and $M$ , on the side of $AB$ opposite $O$ . So $MP = OP - OM = 10 - 6 = 4$ . Choice A (2) comes from subtracting incorrectly. Choice C (6) is $OM$ . Choice D (8) is $AM$ .
A circle has circumference $18\pi$ inches. What is the area of the circle, in square inches?
Answer: $81\pi$ The question asks for the area of a circle with circumference 18π inches. The circumference formula is C = 2πr, so solve for r: 18π = 2πr gives r = 9. Then, area is πr² = π × 81 = 81π square inches. This involves first finding radius from circumference before applying the area formula. A common error is confusing circumference with area or using diameter instead, like mistakenly setting 18π = πd and then area π(d/2)² incorrectly. Always solve for radius explicitly, as it's key to the area formula. In tests, remember that area uses r², so small errors in r amplify in the result.
In the coordinate plane, a circle has equation $(x-3)^2+(y+2)^2=49$ . Which of the following gives the center and radius of the circle? (Be careful about the signs inside the parentheses.)
Answer: Center $(3,-2)$ , $r=7$ This question asks for the center and radius of a circle given its equation in standard form. The standard form of a circle's equation is $(x - h)^2$ + $(y - k)^2$ = $r^2$ , where (h, k) is the center and r is the radius. For the given equation $(x - 3)^2$ + $(y + 2)^2$ = 49, compare it to the standard form to identify h = 3 and k = -2, so the center is (3, -2). Since the right side is 49, which is $r^2$ , take the square root to find r = 7, remembering that radius is the positive square root. A common error is misreading the signs inside the parentheses, such as confusing y + 2 with y - 2, which would incorrectly give k = 2. Another mistake is using 49 as the radius instead of 7, but remember radius is sqrt( $r^2$ ) and must be distinguished from diameter, which is 2r = 14 here. As a test-taking strategy, always double-check the signs and confirm r by taking the square root of the constant term.
A pizza has radius $7$ in. A slice is cut with a central angle of $45^\circ$ . What is the area of the slice, in square inches? (Use $\pi$ in your answer.)
Answer: $\tfrac{49\pi}{8}\text{ in}^2$ This question asks for the area of a pizza slice with a 45° central angle from a pizza with radius 7 inches. The formula for sector area is $A = \frac{\theta}{360^\circ} \times \pi r^2$ . Substituting: $A = \frac{45^\circ}{360^\circ} \times \pi (7)^2 = \frac{1}{8} \times 49\pi = \frac{49\pi}{8}$ square inches. Common mistakes include using diameter instead of radius or incorrectly simplifying the fraction 45/360 to something other than 1/8. When dealing with pizza or pie problems, visualize the fraction of the whole circle.
A circle has diameter $18$ inches. What is the area of the circle, in square inches?
Answer: $81\pi$ in $^2$ This question asks for the area of a circle given its diameter. The area formula is A = πr², where r is the radius. Since diameter = 18 inches, the radius = 18/2 = 9 inches. Substituting into the formula: A = π(9)² = 81π square inches. A critical error is using diameter directly in the formula instead of converting to radius first, which would incorrectly give 324π. Always convert diameter to radius by dividing by 2 before using area or circumference formulas.
A circle is centered at $(2,-1)$ and passes through the point $(8,3)$ . Which equation represents the circle?
Answer: $(x-2)^2+(y+1)^2=52$ This question asks for the equation of a circle given its center and a point it passes through. The center is (2,-1) and it passes through (8,3), so we need to find the radius using the distance formula: r² = (8-2)² + (3-(-1))² = 6² + 4² = 36 + 16 = 52. The equation is therefore (x-2)² + (y+1)² = 52. Remember that in standard form, we use (y-k)², so with k = -1, this becomes (y+1)². A common error is miscalculating the distance or mishandling the signs in the standard form.
Refer to the figure. In circle $P$ , $\angle ARB$ is an inscribed angle that intercepts $\widehat{AB}$ . If $m\angle ARB=35^{\circ}$ , what is $m\widehat{AB}$ ?
Answer: $70^{\circ}$ An inscribed angle measures half its intercepted arc, so the arc measure is $2(35^{\circ})=70^{\circ}$ . C is correct. A gives the angle itself, B mistakenly adds 20°, and D incorrectly doubles twice.
A circular pizza has diameter $14$ inches. If a person eats one slice that is a $45^\circ$ sector of the pizza, what is the area of that slice in terms of $\pi$ (in square inches)?
Answer: $\frac{49\pi}{8}$ This question asks for the area of a 45-degree sector of a pizza with a diameter of 14 inches. The formula for the area of a sector is (theta/360) * π $r^2$ , where theta is the angle and r is the radius. First, note the diameter is 14 inches, so radius r = 7 inches, emphasizing the distinction: radius is half the diameter. Substitute theta = 45 and r = 7: (45/360) π 49 = (1/8) 49π = 49π/8 square inches. A common error is using diameter in the area formula instead of radius, which would give π $(14)^2$ /8 = 196π/8, incorrect because area uses $r^2$ . Another mistake is confusing sector area with arc length, but here we need area. As a strategy, always halve the diameter to get radius before plugging into π $r^2$ .
Refer to the figure. Radius $\overline{OT}$ is drawn to tangent $\overline{TS}$ at point $T$ . If $OT=12$ and $OS=20$ , what is the length of $TS$ ?
Answer: $16$ A radius drawn to a tangent is perpendicular to the tangent at the point of tangency, so triangle $OTS$ is a right triangle with the right angle at $T$ . Using the Pythagorean theorem: $OT^2 + TS^2 = OS^2$ . Substituting the given values: $12^2 + TS^2 = 20^2$ , so $144 + TS^2 = 400$ . Therefore, $TS^2 = 256$ and $TS = 16$ .

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