Question

How do you find the coordinates of the center of a circle whose equation is `x^2-14x+y^2+6y+54=0`?

Answer 1

Complete squares and rearrange into the standard form of the equation of a circle to find centre `(h, k) = (7, -3)` and radius `2`

Explanation:

Complete the squares for both `x` and `y`:

`0 = x^2-14x+y^2+6y+54`

`= (x^2-14x+49) + (y^2+6y+9) + (54-49-9)`

`= (x-7)^2+(y+3)^2-4`

Add `4` to both ends and slightly re-express to get:

`(x-7)^2 + (y-(-3))^2 = 2^2`

This is in the standard form of the equation of a circle:

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

with centre `(h, k) = (7, -3)` and radius `r=2`.

graph{(x^2-14x+y^2+6y+54)((x-7)^2+(y+3)^2-0.01) = 0 [-6.79, 13.21, -6.16, 3.84]}