Chapter 9.2, Problem 3WE

To determine

To calculate: The equation of the circle.

Answer to Problem 3WE

The required equation of the circle is $x^2+y^2-4x+10y+35=0$ .

Explanation of Solution

Given information: The center and the radius of the circle are $\left(2,-5\right)$ and $8$ .

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 , $\left(-g,-f\right)=\left(2,-5\right)$

  $\begin{array}{cc}-g=2& -f=-5\\ \therefore g=-2& \therefore f=5\end{array}$

Now, $r=\sqrt{g^2+f^2-c}=8$

Squaring on both sides,

  $\begin{array}{l}{g}^2+f^2-c=8^2\\ g^2+f^2-c=64\\ {\left(-2\right)}^2+{\left(5\right)}^2-c=64\left\{\because \left(g=-2\right),\left(f=5\right)\right\}\\ \therefore c=64-4-25\\ \therefore c=35\end{array}$

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

  $\begin{array}{l}{x}^2+y^2+2gx+2fy+c=0\\ x^2+y^2+2\times \left(-2\right)x+2\left(5\right)\times y+35=0\\ x^2+y^2-4x+10y+35=0\end{array}$

Therefore, the required equation of the circle is $x^2+y^2-4x+10y+35=0$ .