Chapter 8, Problem 2CR

To determine

The equation of the circle with centre at the point $(-5,1)$ and radius is $3$ and graph the circle.

Answer to Problem 2CR

Solution:

The equation of the circle with centre $(-5,1)$ and radius $3$ is ${\left(x+5\right)}^2+{\left(y-1\right)}^2=9$

To graph the equation, find some points satisfying the above circle equation.

$\begin{array}{ccccc}x& -2& -5& -8& -5\\ y& 1& 4& 1& -2\\ \left(x,y\right)& \left(-2,1\right)& \left(-5,4\right)& \left(-8,1\right)& \left(-5,-2\right)\end{array}$

Draw a circle using the above points

The graph of circle ${\left(x+5\right)}^2+{\left(y-1\right)}^2=9$ is

Explanation of Solution

Given information:

Center of the circle is $(-5,1)$ and radius is $3$

Explanation:

The standard equation of circle with centre $(h,k)$ and radius $r$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Substitute the value of $(h,k)$ as $(-5,1)$ and $r=3$ in the above standard equation.

$\Rightarrow {\left(x-(-5)\right)}^2+{\left(y-1\right)}^2=3^2$

$\Rightarrow {\left(x+5\right)}^2+{\left(y-1\right)}^2=9$

Therefore, the equation of the circle with centre $(-5,1)$ and radius $3$ is ${\left(x+5\right)}^2+{\left(y-1\right)}^2=9$

The graph of ${\left(x+5\right)}^2+{\left(y-1\right)}^2=9$ is