Chapter 9.2, Problem 4WE

To determine

To calculate: The equation of the circle.

Answer to Problem 4WE

The required equation of the circle is $x^2+y^2+6x-2y+15=0$ .

Explanation of Solution

Given information: The center and the radius of the circle are $\left(-3,1\right)$ and $5$ .

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 , $\left(-g,-f\right)=\left(-3,1\right)$

  $\begin{array}{cc}-g=-3& -f=1\\ \therefore g=3& \therefore f=-1\end{array}$

Now, $r=\sqrt{g^2+f^2-c}=5$

Squaring on both sides,

  $\begin{array}{l}{g}^2+f^2-c=5^2\\ g^2+f^2-c=25\\ {\left(3\right)}^2+{\left(-1\right)}^2-c=25\left\{\because \left(g=3\right),\left(f=-1\right)\right\}\\ \therefore c=25-9-1\\ \therefore c=15\end{array}$

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

  $\begin{array}{l}{x}^2+y^2+2gx+2fy+c=0\\ x^2+y^2+2\times \left(3\right)x+2\left(-1\right)\times y+15=0\\ x^2+y^2+6x-2y+15=0\end{array}$

Therefore, the required equation of the circle is $x^2+y^2+6x-2y+15=0$ .