Question

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

Answer 1

(x + 7^)2 + (y − 2)^2 = 81

Explanation:

(x2 + 14x + 49) + (y2 − 4y + 4) − 28 = 49 + 4

(x + 7)2 + (y − 2)2 − 28 = 49 + 4

(x + 7)2 + (y − 2)2 = 49 + 4 + 28 = 81.

Answer 2

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`