Math 2 Quiz: Circle Equations

11 questions · exam conditions

0:00

Circle EquationsQuestion 1 of 11

Points $M(2, 6)$ and $N(8, 2)$ are endpoints of a diameter of a circle. Point $P(a, b)$ lies on this circle such that the area of triangle $MPN$ is $12$ . How many possible positions are there for point $P$ ?

No such point

P

exists on the circle

Exactly one such point

P

exists on the circle

Exactly four such points

P

exist on the circle

Exactly two such points

P

exist on the circle

← Back to quizzes

Math 2 Quiz

Math 2 Quiz: Circle Equations

Practice Circle Equations in Math 2 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 Circle Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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

No such point $P$ exists on the circle
Exactly one such point $P$ exists on the circle
Exactly four such points $P$ exist on the circle
Exactly two such points $P$ exist on the circle (correct answer)

Explanation: When you encounter a problem involving a circle with given diameter endpoints and need to find points with specific geometric properties, start by using the key circle theorems and coordinate geometry principles. Since

M(2, 6)

and

N(8, 2)

are diameter endpoints, the center is at their midpoint:

(\frac{2+8}{2}, \frac{6+2}{2}) = (5, 4)

. The radius is half the distance

MN

r = \frac{1}{2}\sqrt{(8-2)^2 + (2-6)^2} = \frac{1}{2}\sqrt{52} = \sqrt{13}

. For any point

P

on the circle, the area of triangle

MPN

equals

\frac{1}{2} \times \text{base} \times \text{height}

. Using

MN

as the base, we have base length

\sqrt{52} = 2\sqrt{13}

. For area = 12:

12 = \frac{1}{2} \times 2\sqrt{13} \times h

, so the height

h = \frac{12}{\sqrt{13}}

. The height represents the perpendicular distance from

P

to line

MN

. Points at this exact distance from line

MN

form two parallel lines, one on each side of

MN

. Since the required distance

\frac{12}{\sqrt{13}} < \sqrt{13}

(the radius), both parallel lines intersect the circle, giving exactly two intersection points. Choice A is wrong because the required distance is less than the radius, so intersections exist. Choice B incorrectly assumes only one solution, missing that there are two sides to line

MN

. Choice C suggests four points, which would require two distances, but we have only one specific area requirement. Remember: when finding points on a circle at a fixed distance from a chord, there are typically two solutions—one on each side of the chord.

Question 2

A circle is inscribed in the square with vertices at $(0,0)$ , $(6,0)$ , $(6,6)$ , and $(0,6)$ . Point $Q$ is chosen randomly on the circle. What is the probability that both coordinates of $Q$ are greater than $3$ ?

$\frac{1}{8}$
$\frac{1}{6}$
$\frac{1}{3}$
$\frac{1}{4}$ (correct answer)

Explanation: When you encounter probability questions involving geometric regions, you need to find the ratio of favorable area to total area. First, let's establish the setup. The square has side length 6, so the inscribed circle has radius 3 and center at

(3,3)

. The circle's equation is

(x-3)^2 + (y-3)^2 = 9

. You want the probability that both coordinates of point

Q

are greater than 3. This means

Q

must lie in the region where

x > 3

and

y > 3

, which corresponds to the first quadrant relative to the circle's center. Since the circle is perfectly symmetric about both the vertical line

x = 3

and horizontal line

y = 3

, this favorable region represents exactly one-quarter of the circle. The probability is therefore

\frac{\text{area of quarter-circle}}{\text{area of full circle}} = \frac{1}{4}

, making D correct. Let's examine why the other answers are wrong. Choice A (

\frac{1}{8}

) might tempt you if you mistakenly thought about the intersection of

x > 3

and

y > 3

as being more restrictive than it actually is. Choice B (

\frac{1}{6}

) doesn't correspond to any natural geometric division of the circle. Choice C (

\frac{1}{3}

) could result from incorrectly dividing the circle into three parts instead of recognizing the four-way symmetry. Remember: inscribed circles in squares create natural symmetries. When dealing with coordinate conditions like "both greater than the center," you're typically looking at one of four equal quadrants, giving probability

\frac{1}{4}

Question 3

Consider the family of circles with equation $x^2 + y^2 - 4x + 2y + c = 0$ , where $c$ is a parameter. For what range of values of $c$ do these equations represent actual circles (not points or empty sets)?

$c < 5$ (correct answer)
$c < 3$
$c \leq 5$
$c \leq 3$

Explanation: To determine when the equation represents a circle, complete the square:

x^2 - 4x + y^2 + 2y + c = 0

. Adding and subtracting the necessary constants:

(x^2 - 4x + 4) + (y^2 + 2y + 1) + c - 4 - 1 = 0

, which gives

(x-2)^2 + (y+1)^2 = 5 - c

. For this to represent a circle (not a point or empty set), we need the right side to be positive:

5 - c > 0

, so

c < 5

. When

c = 5

, the equation represents a single point (degenerate circle). When

c > 5

, the equation has no real solutions (empty set). Choice C and D incorrectly include the boundary case

c = 5

, while choice B uses the wrong boundary value.

Question 4

A circle has equation $(x-3)^2 + (y+1)^2 = 25$ . Line $\ell$ has equation $4x - 3y = k$ for some constant $k$ . For how many values of $k$ does line $\ell$ intersect the circle at exactly one point?

No values of $k$ result in exactly one intersection
Exactly one value of $k$ results in exactly one intersection
Exactly two values of $k$ result in exactly one intersection (correct answer)
Infinitely many values of $k$ result in exactly one intersection

Explanation: A line intersects a circle at exactly one point when the line is tangent to the circle. This occurs when the distance from the circle's center to the line equals the radius. The circle has center

(3, -1)

and radius

5

. The distance from point

(3, -1)

to line

4x - 3y - k = 0

\frac{|4(3) - 3(-1) - k|}{\sqrt{4^2 + (-3)^2}} = \frac{|12 + 3 - k|}{5} = \frac{|15 - k|}{5}

. For tangency, this distance must equal the radius:

\frac{|15 - k|}{5} = 5

, so

|15 - k| = 25

. This gives

15 - k = 25

15 - k = -25

, so

k = -10

k = 40

. Therefore, there are exactly two values of

k

that result in exactly one intersection point.

Question 5

Circle $C_1$ has center $(2, -1)$ and radius $5$ . Circle $C_2$ has equation $(x-2)^2 + (y+1)^2 = 9$ . Point $M$ lies on both circles. What is the distance from $M$ to the center of $C_1$ ?

$3$
$4$
$5$ (correct answer)
$8$

Explanation: Both circles have the same center

(2, -1)

. Circle

C_1

has radius

5

and circle

C_2

has radius

3

. Since

M

lies on

C_1

, the distance from

M

to the center must equal the radius of

C_1

, which is

5

. Choice A gives the radius of

C_2

. Choice B is the difference of the radii. Choice D is the sum of the radii.

Question 6

Circle $A$ has equation $(x-3)^2 + (y+2)^2 = 16$ and circle $B$ has equation $(x+1)^2 + (y-1)^2 = 9$ . Point $T$ is equidistant from the centers of both circles. If $T$ lies on circle $A$ , what is the $y$ -coordinate of $T$ ?

$-2$
$\frac{1}{2}$
$1$
$2$ (correct answer)

Explanation: Circle

A

has center

(3, -2)

and circle

B

has center

(-1, 1)

. Point

T

equidistant from both centers lies on the perpendicular bisector of the segment connecting the centers. The midpoint is

(1, -\frac{1}{2})

and the slope between centers is

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

. The perpendicular bisector has slope

\frac{4}{3}

and equation

y + \frac{1}{2} = \frac{4}{3}(x-1)

. Since

T

is on circle

A

(x-3)^2 + (y+2)^2 = 16

. Solving simultaneously gives

T = (3, 2)

T = (-1, -6)

. Only

(3, 2)

gives

y = 2

Question 7

Two circles have equations $x^2 + y^2 - 6x + 4y - 12 = 0$ and $x^2 + y^2 - 10x - 6y + 18 = 0$ . At how many points do these circles intersect?

$0$
$1$
$2$ (correct answer)
Cannot be determined from the given information

Explanation: Convert to standard form. Circle 1:

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

with center

(3, -2)

and radius

5

. Circle 2:

(x-5)^2 + (y-3)^2 = 16

with center

(5, 3)

and radius

4

. The distance between centers is

\sqrt{(5-3)^2 + (3-(-2))^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.39

. Since

|5-4| = 1 < \sqrt{29} < 5+4 = 9

, the circles intersect at exactly

2

points. Choice A would occur if circles don't intersect. Choice B would occur if circles are tangent. Choice D suggests insufficient information.

Question 8

A circle has equation $x^2 + y^2 - 8x + 6y - 11 = 0$ . Point $P$ is on the circle such that the $x$ -coordinate of $P$ is twice its $y$ -coordinate. How many such points $P$ exist?

No such points exist on the circle
Exactly one such point exists on the circle
Exactly two such points exist on the circle (correct answer)
Infinitely many such points exist on the circle

Explanation: First, rewrite the circle equation in standard form by completing the square:

x^2 - 8x + y^2 + 6y = 11

. Adding

(8/2)^2 = 16

and

(6/2)^2 = 9

(x-4)^2 + (y+3)^2 = 11 + 16 + 9 = 36

. The circle has center

(4, -3)

and radius

6

. For point

P(x,y)

where

x = 2y

, substitute into the circle equation:

(2y-4)^2 + (y+3)^2 = 36

. Expanding:

4y^2 - 16y + 16 + y^2 + 6y + 9 = 36

, so

5y^2 - 10y + 25 = 36

, giving

5y^2 - 10y - 11 = 0

. Using the quadratic formula:

y = \frac{10 \pm \sqrt{100 + 220}}{10} = \frac{10 \pm \sqrt{320}}{10} = \frac{10 \pm 8\sqrt{5}}{10} = 1 \pm \frac{4\sqrt{5}}{5}

. Since the discriminant is positive, there are exactly two real solutions, meaning exactly two points on the circle satisfy the condition.

Question 9

A circle has the property that it passes through points $(-2, 1)$ , $(4, 1)$ , and $(1, 4)$ . If this circle also passes through point $(1, y_0)$ for some value $y_0 \neq 4$ , what is $y_0$ ?

$-3$
$-2$ (correct answer)
$-1$
$0$

Explanation: First, find the equation of the circle passing through

(-2,1)

(4,1)

, and

(1,4)

. Since the first two points have the same

y

-coordinate, the center lies on the perpendicular bisector of the segment connecting them, which is the vertical line

x = \frac{-2+4}{2} = 1

. So the center is at

(1, k)

for some

k

. Using the fact that the center is equidistant from all three points: distance from

(1,k)

(-2,1)

equals distance from

(1,k)

(1,4)

. We have

\sqrt{(-2-1)^2 + (1-k)^2} = \sqrt{(1-1)^2 + (4-k)^2}

, so

\sqrt{9 + (1-k)^2} = |4-k|

. Squaring both sides:

9 + (1-k)^2 = (4-k)^2

, so

9 + 1 - 2k + k^2 = 16 - 8k + k^2

. Simplifying:

10 - 2k = 16 - 8k

, so

6k = 6

and

k = 1

. The center is

(1,1)

and the radius is

\sqrt{(1-1)^2 + (4-1)^2} = 3

. The circle equation is

(x-1)^2 + (y-1)^2 = 9

. For point

(1, y_0)

to be on this circle:

(1-1)^2 + (y_0-1)^2 = 9

, so

(y_0-1)^2 = 9

, giving

y_0 - 1 = \pm 3

. Thus

y_0 = 4

y_0 = -2

. Since we're told

y_0 \neq 4

, we have

y_0 = -2

Question 10

The equation $x^2 + y^2 + Dx + Ey + F = 0$ represents a circle with center $(-3, 4)$ and radius $7$ . What is the value of $D + E + F$ ?

$-24$
$-26$ (correct answer)
$24$
$26$

Explanation: The standard form is

(x+3)^2 + (y-4)^2 = 49

. Expanding:

x^2 + 6x + 9 + y^2 - 8y + 16 = 49

, so

x^2 + y^2 + 6x - 8y - 24 = 0

. Thus

D = 6

E = -8

F = -24

, and

D + E + F = 6 + (-8) + (-24) = -26

. Choice A omits

D + E

. Choice C gets the sign wrong. Choice D gets both the calculation and sign wrong.

Question 11

The equation $x^2 + y^2 + 6x - 8y + k = 0$ represents a circle that passes through the origin. What is the area of this circle?

$25\pi$ (correct answer)
$50\pi$
$\pi$
$100\pi$

Explanation: Since the circle passes through the origin

(0,0)

, substituting gives

0 + 0 + 0 - 0 + k = 0

, so

k = 0

. The equation becomes

x^2 + y^2 + 6x - 8y = 0

. Completing the square:

(x+3)^2 - 9 + (y-4)^2 - 16 = 0

, so

(x+3)^2 + (y-4)^2 = 25

. The radius is

5

and the area is

\pi r^2 = 25\pi

. Choice B doubles the area. Choice C uses radius

1

. Choice D uses diameter squared instead of radius squared.