Chapter 9.2, Problem 7WE

To determine

To calculate: Find the equation of the circle.

Answer to Problem 7WE

The required equation of the circle is $x^2+y^2-12x-2y+35=0$ .

Explanation of Solution

Given information: The circle has coordinates of the center of $\left(6,1\right)$ and the radius of $\sqrt{2}$ .

Formula used: The distance formula: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation:

As we know that the center of a circle $=\left(h,k\right)$

Now, substitute the value of the coordinates of the center and the radius of the circle into ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Hence,

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-6\right)}^2+{\left(y-1\right)}^2={\left(\sqrt{2}\right)}^2\left[\because \left(h,k\right)=\left(6,1\right)\right]\\ \left(x^2+36-12x\right)+\left(y^2+1-2y\right)=2\\ x^2+y^2-12x-2y+36+1-2=0\\ x^2+y^2-12x-2y+35=0\end{array}$

Therefore, the required equation of the circle is $x^2+y^2-12x-2y+35=0$ .