Chapter 9.2, Problem 8WE

To determine

To calculate: Find the equation of the circle.

Answer to Problem 8WE

The required equation of the circle is $x^2+y^2+10x-6y+33.98=0$ .

Explanation of Solution

Given information: The circle has coordinates of the center of $\left(-5,3\right)$ and the radius of $\frac{1}{6}$ .

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,

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-\left(-5\right)\right)}^2+{\left(y-3\right)}^2={\left(\frac{1}{6}\right)}^2\left[\because \left(h,k\right)=\left(-5,3\right)\right]\\ {\left(x+5\right)}^2+{\left(y-3\right)}^2={\left(\frac{1}{6}\right)}^2\\ x^2+25+10x+y^2+9-6y=\frac{1}{36}\\ 36x^2+36y^2+360x-216y+1224=1\\ x^2+y^2+10x-6y+33.98=0\end{array}$

Therefore, the required equation of the circle is $x^2+y^2+10x-6y+33.98=0$ .