Question

What is the center and radius of the circle with the given equation x^2 + y^2 + 14x - 12y +69 = 0

Answer 1

Answer:

Center is (7,-6)

Our radius is 4

Step-by-step explanation:

Subtract 69 to both sides

Complete the square of the x terms and y terms.

${x}^{2} + 14x$

${y}^{2} - 12y$

(To complete the square divide the coefficient by 2 and square it).

$(\frac{14}{2} ) {}^{2} = 49$

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

Add those terms to the full equation. Add 36 and 49 to the right side as well

${x}^{2} + 14x + 49 + {y}^{2} - 12y + 36 = 16$

Simplify the x and y terms into a binomial.

$(x + 7) {}^{2} + (y - 6) {}^{2} = 16$

Set the left side equal to zero separately to find the center and take the square root of the right side to find the radius.

$x + 7 = 0$

$x = - 7$

$y - 6 = 0$

$y = 6$

$16 = {x}^{2}$

$4 = x$

So our center is (-7,6)

Our radius is 4.