Question

What is the Center and radius of 6x + 37 = 20y - x^2 -y^2

Answer 1

We are given the equation of the circle

`6x+37=20y-x^2-y^2`

We will first rewrite the equation

`\begin{gathered} x^2+y^2+6x-20y+37=0 \\ \text{Collect like terms} \\ x^2+6x+y^2-20y+37=0 \\ (x^2+6x)+(y^2-20y)=-37 \end{gathered}`

Now we complete to perfect square in the bracket

Following the same process, we have

`\begin{gathered} y^2-20y \\ so\text{ we will have} \\ (y-10)^2-100 \end{gathered}`

Now, from

`\begin{gathered} (x^2+6x)+(y^2-20y)=-37 \\ (x+3)^2-9+(y-10)^2-100=-37 \\ (x+3)^2+(y-10)^2=-37+9+100 \\ (x+3)^2+(y-10)^2=72 \\ \end{gathered}`

Comparing with the equation of a circle

`(x-a)^2+(y-b)^2=r^2`

Therefore,

`\begin{gathered} Centre=(-3,10) \\ Radius^2=72 \\ Radius=\sqrt[]{72} \\ Radius=\sqrt[]{36*2} \\ Radius=6\sqrt[]{2} \end{gathered}`