Question

What is the equation of a circle with its center at (−6,−3) and a radius of 12?

Answer 1

x² + y² + 12x +6y - 99 = 0

Explanation:

Given :

• Centre = (-6,-3)
• Radius = 12 ,

Using the Standard equation of circle ,

• ( x - h)² + (y-k)² = r²
• ( x +6)² + (y+3)² = 12²
• x² + 36 + 12x + y² + 9 + 6y = 144
• x² + y² + 12x + 6y +45-144 = 0
• x² + y² + 12x +6y - 99 = 0

Answer 2

(x+6)^2 + ( y +3)^2 = 144

Explanation:

The equation of a circle is given by

(x-h)^2 +(y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x- -6)^2 + ( y - -3)^2 = 12^2

(x+6)^2 + ( y +3)^2 = 144