In the standard (x, y) coordinate plane, a circle is tangent to the horizontal line y = 7.… — ACT

In the standard $(x, y)$ coordinate plane, a circle is tangent to the horizontal line $y = 7$ . The equation of the circle is $x^2 + y^2 - 10x + 4y + c = 0$ , where $c$ is a constant. What is the value of $c$ ?

Answer: -52

Answer

The value of

c

-52

Completing the square gives

(x - 5)^2 + (y + 2)^2 = 29 - c

. The center is

(5, -2)

. The distance from

(5, -2)

to the horizontal line

y = 7

|7 - (-2)| = 9

, which is the radius

r

. Squaring the radius gives

r^2 = 81

. Setting

29 - c = 81

gives

c = -52

Step-by-Step Solution

Complete the square for

x

and

y

terms.

(x - 5)^2 + (y + 2)^2 = 29 - c

Writing the equation in standard form

(x - h)^2 + (y - k)^2 = r^2

reveals the center

(h, k)

and the expression for

r^2

Calculate the radius using the distance from the center to the tangent line.

Radius

r = 9

The center is

(5, -2)

and the tangent line is

y = 7

. The distance between

y = -2

and

y = 7

|7 - (-2)| = 9

Set

r^2 = 81

and solve for

c

c = -52

Equating the radius squared expression

29 - c

81

yields

29 - c = 81 \implies c = -52

Key Concept

Converting circle equations to standard form and using distance to tangent lines to determine radius.

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