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

Jul 27, 2017

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

#### Explanation:

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

${x}^{2} + 18 x + {y}^{2} + 56 = 0$

Then let's write in standard form ${x}^{2} + {y}^{2} + a x + b y + c = 0$:

${x}^{2} + {y}^{2} + 18 x + 56 = 0$

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)

$r = \frac{1}{2} \sqrt{{a}^{2} + {b}^{2} - 4 c}$

Then we will get:

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

$r = \frac{1}{2} \sqrt{{18}^{2} + 0 - 4 \cdot 56} = \frac{1}{2} \sqrt{100} = \frac{1}{2} \cdot 10 = 5$

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