Question

How do you write the equation `-2x^2-36x-2y^2-112=0` in standard form and find the center and radius?

Answer 1

`x^2+y^2+18x+56=0`
`C=(-9;0)`
`r=5`

Explanation:

First, let's divide all terms of the equation by `-2` and get:

`x^2+18x+y^2+56=0`

Then let's write in standard form `x^2+y^2+ax+by+c=0`:

`x^2+y^2+18x+56=0`

where

`a=18`
`b=0`
`c=56`

Now we can get the center and radius by applying the following formulas:

`C=(-a/2;-b/2)`

`r=1/2sqrt(a^2+b^2-4c)`

Then we will get:

`C=(-cancel18^9/cancel2;0)=(-9;0)`

`r=1/2sqrt(18^2+0-4*56)=1/2sqrt100=1/2*10=5`

graph{x^2+y^2+18x+56=0 [-18, 3, -6, 6]}