Question

What is the center of the circle given by the equation x^2+y^2-14y-15=0

Answer 1

(0, 7)

Explanation:

The equation of a circle in the standard form:

`(a-h)^2+(y-k)^2=r^2`

(h, k) - center

r - radius

We have the equation:

`x^2+y^2-12y-15=0`

Convert into a standard form using

`(a-b)^2=a^2-2ab+b^2\qquad(*)`

`x^2+\underbrace{y^2-2(y)(7)+7^2}_((*))-7^2-15=0\\\\(x-0)^2+(y-7)^2-49-15=0`

`(x-0)^2+(y-7)^2-64=0` add 64 to both sides

`(x-0)^2+(y-7)^2=64`

The center (0, 7)

The radius: r = √64 = 8