In the xy-plane, the circle defined by the equation x^2 + y^2 - 10x + 12y = k has a radius… — SAT

In the $xy$ -plane, the circle defined by the equation $x^2 + y^2 - 10x + 12y = k$ has a radius of $9$ , where $k$ is a constant. What is the value of $k$ ?

A
-52
B
70
20Answer
D
142

Answer

To find the value of

k

, we first rewrite the given equation in the standard form of a circle's equation,

(x - h)^2 + (y - k_{center})^2 = r^2

. Starting with

x^2 + y^2 - 10x + 12y = k

, we complete the square for the

x

-terms by adding

(\frac{-10}{2})^2 = 25

to both sides, and for the

y

-terms by adding

(\frac{12}{2})^2 = 36

to both sides. This yields

(x^2 - 10x + 25) + (y^2 + 12y + 36) = k + 25 + 36

, which simplifies to

(x - 5)^2 + (y + 6)^2 = k + 61

. In this standard form, the right-hand side represents the square of the radius, so

r^2 = k + 61

. Given that the radius is

9

, we have

r^2 = 9^2 = 81

. Setting

k + 61 = 81

and subtracting

61

from both sides gives

k = 20

Step-by-Step Solution

Write the given equation of the circle.

x^2 + y^2 - 10x + 12y = k

To establish the starting equation before completing the square.

Complete the square for the

x

and

y

terms by adding

(\frac{-10}{2})^2 = 25

and

(\frac{12}{2})^2 = 36

to both sides.

(x^2 - 10x + 25) + (y^2 + 12y + 36) = k + 25 + 36

To express the quadratic expressions as perfect squares.

Rewrite the equation in standard form.

(x - 5)^2 + (y + 6)^2 = k + 61

To match the standard equation of a circle,

(x-h)^2 + (y-k_{center})^2 = r^2

, where the right-hand side represents the square of the radius.

Equate the constant term on the right-hand side to

r^2

using the given radius

r = 9

, and solve for

k

k + 61 = 9^2 \implies k + 61 = 81 \implies k = 20

To calculate the value of the constant

k

that satisfies the radius requirement.

Key Concept

Equations of Circles in the Coordinate Plane

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