Chapter 12, Problem 14STP

To determine

To find: Radius of the circle.

Answer to Problem 14STP

Radius of given circle equation is $r=2$

Explanation of Solution

Given information:

  $x^2+y^2+8x+8y+28=0$

Concept:

If circle equation is in the form

  $x^2+y^2+2gx+2hy+c=0$

Then radius of circle is

  $r=\sqrt{h^2+k^2-c}$

Given

  $x^2+y^2+8x+8y+28=0$

On comparing with general equation

We get

  $\begin{array}{l}2g=8\\ 2h=8\\ c=28\end{array}$

On simplifying

  $\begin{array}{l}2g=8\\ \Rightarrow \frac{2g}{2}=\frac{8}{2}\\ \Rightarrow g=4\\ 2h=8\\ \Rightarrow \frac{2h}{2}=\frac{8}{2}\\ \Rightarrow h=4\\ c=28\end{array}$

We know

Radius $=r=\sqrt{h^2+k^2-c}$

On substituting values

We get

  $\begin{array}{l}\Rightarrow r=\sqrt{h^2+k^2-c}\\ \Rightarrow r=\sqrt{4^2+4^2-28}\\ \Rightarrow r=\sqrt{16+16-28}\\ \Rightarrow r=\sqrt{32-28}\\ \Rightarrow r=\sqrt{4}\\ \Rightarrow r=2\end{array}$

Therefore,

Radius of given circle equation is $r=2$