Chapter 9.2, Problem 1WE

To determine

To calculate: The equation of the circle.

Answer to Problem 1WE

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

Explanation of Solution

Given information: The center and the radius of the circle are $\left(3,0\right)$ and $3$ .

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 , $\left(-g,-f\right)=\left(3,0\right)$

  $\begin{array}{ccc}-g=3& -f=0& \\ \therefore g=-3& \therefore f=0& \end{array}$

Now, $r=\sqrt{g^2+f^2-c}=3$

Squaring on both sides,

  $\begin{array}{l}{g}^2+f^2-c=3^2\\ g^2+f^2-c=9\\ -3^2+0^2-c=9\left\{\because \left(g=-3\right),\left(f=0\right)\right\}\\ \therefore c=9-9\\ \therefore c=0\end{array}$

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

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

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