What is the center and the radius of a circle with the equation:

x^2 + 6x + y^2 - 8y - 11 = 0.

+1 vote

Given circle equation x ^2 + 6x +y ^2 -8y -11 = 0.

Compare it to circle equation x ^2 + y ^2 + 2g x +2f y +c = 0

c = -11

2= 6

Divide each side by 2.

g  = 3

2f  = -8

Divide each side by 2.

f  = -4

Center = (-g , -) = ( -3 , 4 ).

Center = ( -3 , 4 )

The standard form of the circle equation is ( x - h )2 + ( y - k )2 = r2, where, (h, k) is the center of the circle, and r is the radius.

The equation is x2 + 6x +  y2 - 8y - 11 = 0.

Write the equation in standard form of a circle.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here, x coefficient = 6, so, (half the x coefficient)² = (6/2)2= 9.

Here, y coefficient = - 8, so, (half the y coefficient)² = (- 8/2)2= 16.

Add 9 and 16 to each side.

x2 + 6x + 9 + y2 - 8y + 16 - 11 = 0 + 9 + 16

(x + 3)2 + (y - 4)2 = 11 + 25 = 36

(x - (- 3))2 + (y - 4))2 = 62 .

Compare the equation with standard form of a circle equation.

The center (h, k) is (- 3, 4), and

The radius (r) is 6 units.