Chapter 1.8, Problem 96E

To determine

To calculate: Theequation of a circle whose center is at $\left(-1,5\right)$ and passes through the point $\left(-4,-6\right)$ .

Answer to Problem 96E

Theequation of the circle is ${\left(x+1\right)}^2+{\left(y-5\right)}^2=130$ .

Explanation of Solution

Given information:

The center of a circle is at $\left(-1,5\right)$ and passes through the point $\left(-4,-6\right)$ .

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 conditions that center of a circle is at $\left(-1,5\right)$ and passes through the point $\left(-4,-6\right)$ .

Since the circle passes through the point $\left(-4,-6\right)$ , and center is at $\left(-1,5\right)$ therefore radius of circle is the distance between the points $\left(-4,-6\right)$ and $\left(-1,5\right)$ .

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,

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(-4-\left(-1\right)\right)}^2+{\left(-6-5\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(-11\right)}^2}\\ =\sqrt{9+121}\\ =\sqrt{130}\end{array}$

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 $\left(-1,5\right)$ also radius ris $\sqrt{130}$ .

Here, $h=-1,k=5\text{ and }r=\sqrt{130}$ .

Substitute the values in standard equation of circle,

  $\begin{array}{c}{\left(x-\left(-1\right)\right)}^2+{\left(y-5\right)}^2={\left(\sqrt{130}\right)}^2\\ {\left(x+1\right)}^2+{\left(y-5\right)}^2=130\end{array}$

Thus, the equation of circle is ${\left(x+1\right)}^2+{\left(y-5\right)}^2=130$ .