Question

How do you find the center and radius of the circle given `x^2+y^2+6y=-50-14x`?

Answer 1

Centre (-7,-3) radius `sqrt8`

Explanation:

`(x-a)^2+(y-b)^2=r^2` is the standard form for the equation of a circle with `(a,b)` the centre and radius `r`

Rearrange to this form:
`x^2+14x+y^2+6y=-50`

Complete the square:
`[(x+7)^2-49]+[(y+3)^2-9]=-50`

`(x +7)^2+(y+3)^2=-50+58`

Collect terms and tidy up
`(x +7)^2+(y+3)^2=8`