MATH SOLVE

2 months ago

Q:
# What is the standard form of the equation of the circle x2 -4x + y2 + 6y + 12 = 0?

Accepted Solution

A:

We take the equation

x^2 -4x + y^2 + 6y + 12 = 0

and complete the square for x and y.

x^2 -4x + 4 - 4 + y^2 + 6y + 9 - 9 + 12= 0

(x-2)^2 - 4 + (y+3)^2 - 9 = -12

(x-2)^2+ (y+3)^2 = 1

Therefore the answer is (x-2)^2+ (y+3)^2 = 1

x^2 -4x + y^2 + 6y + 12 = 0

and complete the square for x and y.

x^2 -4x + 4 - 4 + y^2 + 6y + 9 - 9 + 12= 0

(x-2)^2 - 4 + (y+3)^2 - 9 = -12

(x-2)^2+ (y+3)^2 = 1

Therefore the answer is (x-2)^2+ (y+3)^2 = 1