Chapter 9.2, Problem 5WE

To determine

To calculate: The equation of the circle.

Answer to Problem 5WE

The required equation of the circle is $x^2+y^2-144=0$ .

Explanation of Solution

Given information: The center and the radius of the circle are $\left(0,0\right)$ and $12$ .

Formula used: $x^2+y^2+2gx+2fy+c=0$ and $r=\sqrt{g^2+f^2-c}$

Calculation: As we know that $\left(-g,-f\right)$ is the center of a circle and $r=\sqrt{g^2+f^2-c}$ is the radius of a circle.

Hence , $\left(-g,-f\right)=\left(0,0\right)$

  $\begin{array}{cc}-g=0& -f=0\\ \therefore g=0& \therefore f=0\end{array}$

Now, $r=\sqrt{g^2+f^2-c}=12$

Squaring on both sides,

  $\begin{array}{l}{g}^2+f^2-c={12}^2\\ g^2+f^2-c=144\\ {\left(0\right)}^2+{\left(0\right)}^2-c=144\left\{\because \left(g=0\right),\left(f=0\right)\right\}\\ \therefore -c=144\\ \therefore c=-144\end{array}$

Now, substitute the value of $g,fandc$ in the general equation of a circle.

  $\begin{array}{l}{x}^2+y^2+2gx+2fy+c=0\\ x^2+y^2+2\times \left(0\right)x+2\left(0\right)\times y-144=0\\ x^2+y^2-144=0\end{array}$

Therefore, the required equation of the circle is $x^2+y^2-144=0$ .