Chapter 1.8, Problem 95E

To determine

To calculate: Theequation of a circle whose center is at origin and passes through the point $\left(4,7\right)$ .

Answer to Problem 95E

Theequation of the circle is $x^2+y^2=65$ .

Explanation of Solution

Given information:

The center of a circle is at origin and passes through the point $\left(4,7\right)$ .

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 origin and passes through the point $\left(4,7\right)$ .

Since the circle passes through the point $\left(4,7\right)$ , and center is at $\left(0,0\right)$ therefore radius of circle is the distance between the points $\left(4,7\right)$ and $\left(0,0\right)$ .

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,

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(4-0\right)}^2+{\left(7-0\right)}^2}\\ =\sqrt{16+49}\\ =\sqrt{65}\end{array}$

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 $\left(0,0\right)$ also radius ris $\sqrt{65}$ .

Here, $h=0,k=0\text{ and }r=\sqrt{65}$ .

Substitute the values in standard equation of circle,

  $\begin{array}{c}{\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(\sqrt{65}\right)}^2\\ x^2+y^2=65\end{array}$

Thus, the equation of circle is $x^2+y^2=65$ .