Chapter 1.8, Problem 101E

To determine

To calculate: Theequation of a circle from the figure provided below,

  

Answer to Problem 101E

Theequation of the circle is ${\left(x+2\right)}^2+{\left(y-2\right)}^2=4$ .

Explanation of Solution

Given information:

The graph of the circle is provided below,

  

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.

Distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

Consider the provided graph of the circle,

  

It is observed that center of the circle is at $\left(-2,2\right)$ .

Since the circle passes through the point $\left(0,2\right)$ , and center is at $\left(-2,2\right)$ therefore radius of circle is the distance between the points $\left(0,2\right)$ and $\left(-2,2\right)$ .

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Apply it,

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(-2-0\right)}^2+{\left(2-2\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(0\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

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, $\left(h,k\right)$ and $\left(-2,2\right)$ also radius ris $2$ .

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

Substitute the values in standard equation of circle,

  $\begin{array}{c}{\left(x-\left(-2\right)\right)}^2+{\left(y-2\right)}^2=2^2\\ {\left(x+2\right)}^2+{\left(y-2\right)}^2=4\end{array}$

Thus, the equation of circle is ${\left(x+2\right)}^2+{\left(y-2\right)}^2=4$ .