Chapter 1.8, Problem 92E

To determine

To calculate: Thecenter and radius of the circle ${\left(x+1\right)}^2+{\left(y+2\right)}^2=36$ and sketch the graph of the circle.

Answer to Problem 92E

The center of the circle is $\left(-1,-2\right)$ and radius is $6$ units. The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation ${\left(x+1\right)}^2+{\left(y+2\right)}^2=36$ .

Formula used:

The standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Calculation:

Consider the equation ${\left(x+1\right)}^2+{\left(y+2\right)}^2=36$ .

Rewrite the equation as,

  ${\left(x-\left(-1\right)\right)}^2+{\left(y-\left(-2\right)\right)}^2=6^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-\left(-1\right)\right)}^2+{\left(y-\left(-2\right)\right)}^2=6^2$ .

Here, $h=-1,k=-2\text{ and }r=6$ .

Therefore, center of circle is $\left(-1,-2\right)$ and radius is $6$ .

Now, on the Cartesian plane construct a circle with center at $\left(-1,-2\right)$ and radius $6$ units.

The graph obtained is provided below,

  

Thus, the center of the circle is $\left(-1,-2\right)$ and radius is $6$ units.