Math 3 Quiz: Writing Equations From Geometry

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Writing Equations From GeometryQuestion 1 of 9

A line is tangent to the circle $(x - 3)^2 + (y + 2)^2 = 25$ at the point $(7, 1)$ . What is the equation of this tangent line?

4x + 3y = 31

4x - 3y = 25

3x + 4y = 25

3x - 4y = 17

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Math 3 Quiz

Math 3 Quiz: Writing Equations From Geometry

Practice Writing Equations From Geometry in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Writing Equations From Geometry, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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.

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Question 1

A line is tangent to the circle $(x - 3)^2 + (y + 2)^2 = 25$ at the point $(7, 1)$ . What is the equation of this tangent line?

$4x + 3y = 31$ (correct answer)
$4x - 3y = 25$
$3x + 4y = 25$
$3x - 4y = 17$

Explanation: The tangent line at a point on a circle is perpendicular to the radius at that point. The center is

(3, -2)

and the point of tangency is

(7, 1)

. The slope of the radius is

\frac{1-(-2)}{7-3} = \frac{3}{4}

. Therefore, the slope of the tangent line is

-\frac{4}{3}

. Using point-slope form:

y - 1 = -\frac{4}{3}(x - 7)

, which gives

y - 1 = -\frac{4}{3}x + \frac{28}{3}

. Rearranging:

y = -\frac{4}{3}x + \frac{28}{3} + 1 = -\frac{4}{3}x + \frac{31}{3}

. Multiplying by 3:

3y = -4x + 31

, so

4x + 3y = 31

. Choice B uses the slope of the radius instead of its negative reciprocal. Choice C incorrectly switches the coefficients. Choice D has the wrong constant term.

Question 2

In triangle $ABC$ , the perpendicular bisector of side $AB$ passes through point $C(7, 3)$ . If $A(-1, 5)$ and $B(3, -1)$ , what is the equation of the line containing side $AC$ ?

$y = -\frac{1}{4}x + \frac{19}{4}$ (correct answer)
$y = -\frac{1}{4}x + \frac{17}{4}$
$y = \frac{2}{3}x + \frac{17}{3}$
$y = \frac{2}{3}x + \frac{19}{3}$

Explanation: Since the perpendicular bisector of AB passes through C, triangle ABC is isosceles with CA = CB. The perpendicular bisector of AB has midpoint (1, 2) and slope 2/3 (negative reciprocal of AB's slope -3/2). Its equation is y - 2 = (2/3)(x - 1). We can verify C(7, 3) lies on this line: 3 - 2 = (2/3)(7 - 1) = 4, which is true. Now for line AC with A(-1, 5) and C(7, 3): slope = (3-5)/(7-(-1)) = -2/8 = -1/4. Using point-slope form with point A: y - 5 = (-1/4)(x - (-1)) = (-1/4)(x + 1), so y = -x/4 - 1/4 + 5 = -x/4 + 19/4.

Question 3

Circle $C_1$ has equation $(x - 2)^2 + (y + 3)^2 = 16$ and circle $C_2$ has equation $(x + 1)^2 + (y - 1)^2 = 9$ . What is the equation of the line containing the centers of both circles?

$4x + 3y = -1$ (correct answer)
$4x - 3y = 17$
$3x + 4y = -6$
$3x - 4y = 18$

Explanation: The center of circle C₁ is (2, -3) and the center of circle C₂ is (-1, 1). The line through these centers has slope (1 - (-3))/(-1 - 2) = 4/(-3) = -4/3. Using point-slope form with center (2, -3): y - (-3) = (-4/3)(x - 2), so y + 3 = (-4/3)x + 8/3, giving y = (-4/3)x + 8/3 - 3 = (-4/3)x - 1/3. Converting to standard form: 3y = -4x - 1, so 4x + 3y = -1.

Question 4

Line $\ell$ is tangent to the circle $(x - 4)^2 + (y + 1)^2 = 25$ at the point $(8, 2)$ . What is the equation of line $\ell$ ?

$4x + 3y = 38$ (correct answer)
$3x + 4y = 32$
$4x - 3y = 26$
$3x - 4y = 16$

Explanation: A tangent line to a circle is perpendicular to the radius at the point of tangency. The center of the circle is (4, -1) and the point of tangency is (8, 2). The slope of the radius from (4, -1) to (8, 2) is (2-(-1))/(8-4) = 3/4. The tangent line has slope that is the negative reciprocal: -4/3. Using point-slope form with point (8, 2): y - 2 = -4/3(x - 8), which gives y - 2 = -4x/3 + 32/3, so y = -4x/3 + 32/3 + 2 = -4x/3 + 38/3. Multiplying by 3: 3y = -4x + 38, or 4x + 3y = 38.

Question 5

A circle is inscribed in a square with vertices at $(0, 0)$ , $(6, 0)$ , $(6, 6)$ , and $(0, 6)$ . What is the equation of this inscribed circle?

$(x - 3)^2 + (y - 3)^2 = 9$ (correct answer)
$(x - 3)^2 + (y - 3)^2 = 6$
$(x - 3)^2 + (y - 3)^2 = 3$
$(x - 3)^2 + (y - 3)^2 = 36$

Explanation: An inscribed circle touches all four sides of the square. The center of the inscribed circle is at the center of the square, which is the midpoint of the diagonal from (0,0) to (6,6), giving us (3,3). The radius of the inscribed circle is half the side length of the square. The side length is 6, so the radius is 3. Therefore, the equation is (x-3)² + (y-3)² = 3² = 9.

Question 6

The center of a circle lies on the line $2x + y = 10$ , and the circle is tangent to both the $x$ -axis and the line $y = 4$ . What is the equation of this circle?

$(x - 3)^2 + (y - 4)^2 = 4$
$(x - 4)^2 + (y - 2)^2 = 4$ (correct answer)
$(x - 2)^2 + (y - 6)^2 = 4$
$(x - 5)^2 + (y - 0)^2 = 4$

Explanation: Since the circle is tangent to both the

x

-axis and the line

y = 4

, the center must be equidistant from both lines. The distance from a point

(h, k)

to the

x

-axis is

|k|

, and the distance to the line

y = 4

|k - 4|

. For tangency to both:

|k| = |k - 4|

. This gives

k = k - 4

(impossible) or

k = -(k - 4) = -k + 4

, so

2k = 4

and

k = 2

. The radius is

r = 2

. Since the center lies on

2x + y = 10

2h + 2 = 10

, so

h = 4

. Therefore, the center is

(4, 2)

and the equation is

(x - 4)^2 + (y - 2)^2 = 4

. Choice A has the wrong center. Choice C has

k = 6

, which would make the circle intersect the

x

-axis, not be tangent to it. Choice D has the center on the

x

-axis, making it impossible to be tangent to both lines with the same radius.

Question 7

A circle has its center at $(-2, 3)$ and passes through the point $(1, 7)$ . If this circle intersects the $y$ -axis at two points, what is the $y$ -coordinate of the lower intersection point?

$3 - 2\sqrt{6}$
$3 - \sqrt{6}$
$3 - \sqrt{15}$
$3 - \sqrt{21}$ (correct answer)

Explanation: When you encounter a circle problem involving intersection points, you need to find the circle's equation first, then solve for where it meets the given line or axis. Start by finding the radius using the distance formula between the center

(-2, 3)

and the given point

(1, 7)

r = \sqrt{(1-(-2))^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

The circle's equation is

(x + 2)^2 + (y - 3)^2 = 25

. To find where this circle intersects the

y

-axis, substitute

x = 0

(0 + 2)^2 + (y - 3)^2 = 25

4 + (y - 3)^2 = 25

(y - 3)^2 = 21

y - 3 = \pm\sqrt{21}

y = 3 \pm \sqrt{21}

The two intersection points are at

y = 3 + \sqrt{21}

(upper) and

y = 3 - \sqrt{21}

(lower). The lower point is

3 - \sqrt{21}

, which is choice D. Choice A (

3 - 2\sqrt{6}

) likely comes from incorrectly calculating

2\sqrt{6} = \sqrt{24}

instead of

\sqrt{21}

. Choice B (

3 - \sqrt{6}

) might result from forgetting to square the radius or making an arithmetic error. Choice C (

3 - \sqrt{15}

) could come from incorrectly computing

25 - 4 = 15

but missing a step in the calculation. Remember: always verify your radius calculation first, then carefully substitute into the circle equation. Double-check your algebra when solving for intersection points.

Question 8

The line $3x + 4y = 24$ intersects the coordinate axes to form a triangle with the origin. What is the equation of the circle inscribed in this triangle?

$(x - 2)^2 + (y - 2)^2 = 4$ (correct answer)
$(x - 3)^2 + (y - 3)^2 = 9$
$(x - 2)^2 + (y - 3)^2 = 4$
$(x - 3)^2 + (y - 2)^2 = 4$

Explanation: First, find the intercepts. For

x

-intercept: set

y = 0

, so

3x = 24

and

x = 8

. For

y

-intercept: set

x = 0

, so

4y = 24

and

y = 6

. The triangle has vertices at

(0, 0)

(8, 0)

, and

(0, 6)

. For a right triangle with legs of length

a

and

b

and hypotenuse

c

, the inradius is

r = \frac{a + b - c}{2}

. Here,

a = 8

b = 6

, and

c = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10

. So

r = \frac{8 + 6 - 10}{2} = \frac{4}{2} = 2

. The incenter of a right triangle with legs along the axes is at

(r, r) = (2, 2)

. Therefore, the equation is

(x - 2)^2 + (y - 2)^2 = 4

. Choice B has the wrong center and radius. Choices C and D have incorrect centers that don't correspond to the incenter of this triangle.

Question 9

A rectangle has vertices at $A(1, 2)$ , $B(7, 4)$ , $C(6, 7)$ , and $D(0, 5)$ . What is the equation of the circle that passes through all four vertices of this rectangle?

$(x - 3)^2 + (y - 4)^2 = 13$
$(x - 4)^2 + (y - 5)^2 = 10$
$(x - 3.5)^2 + (y - 4.5)^2 = 12.5$ (correct answer)
$(x - 2)^2 + (y - 3)^2 = 18$

Explanation: When a rectangle is inscribed in a circle, the circle passes through all four vertices, and its center is at the intersection of the rectangle's diagonals. Since a rectangle's diagonals bisect each other, you can find the circle's center by finding the midpoint of either diagonal. Let's find the center using diagonal

AC

. The midpoint formula gives us: Center =

\left(\frac{1+6}{2}, \frac{2+7}{2}\right) = (3.5, 4.5)

We can verify this is correct by checking diagonal

BD

: Midpoint =

\left(\frac{7+0}{2}, \frac{4+5}{2}\right) = (3.5, 4.5)

✓ Now we need the radius. Calculate the distance from the center

(3.5, 4.5)

to any vertex, say

A(1, 2)

r^2 = (1-3.5)^2 + (2-4.5)^2 = (-2.5)^2 + (-2.5)^2 = 6.25 + 6.25 = 12.5

The circle equation is

(x - 3.5)^2 + (y - 4.5)^2 = 12.5

, which is choice C. Choice A has center

(3, 4)

and radius squared

13

- this center doesn't match the diagonal midpoint. Choice B uses center

(4, 5)

with radius squared

10

- again, wrong center location. Choice D places the center at

(2, 3)

with radius squared

18

- this center is also incorrect and would place the circle too far from the actual vertices. Strategy tip: For any quadrilateral inscribed in a circle, the center is where the perpendicular bisectors of the sides meet. For rectangles specifically, this simplifies to the midpoint of either diagonal, making the calculation much faster.