Question

What is the standard form of this equation of a circle? x2 + y2 + 14x - 4y - 28 = 0

Answer 1

We complete the square.

$x^2 + y^2 + 14x - 4y - 28 = 0$

$x^2+14x + y^2 - 4y = 28$

$x^2+14x + (14/2)^2 + y^2 - 4y + (4/2)^2 = 28 + 49 + 4 = 81$

So the answer is:

$(x+7)^2 + (y-2)^2 = 9^2$