# Theequation of a circle whose center is at origin and passes through the point ( 4 , 7 ) .

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.8, Problem 95E
To determine

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

Expert Solution

Theequation of the circle is x2+y2=65 .

### Explanation of Solution

Given information:

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

Formula used:

The standard form of the equation of the circle is (xh)2+(yk)2=r2 , where (h,k) denote the center of the circle and r denote the radius.

Distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .

Calculation:

Consider the provided conditions that center of a circle is at origin and passes through the point (4,7) .

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

Recall that the distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .

Apply it,

d(A,B)=(40)2+(70)2=16+49=65

Recall that the standard form of the equation of the circle is (xh)2+(yk)2=r2 , where (h,k) denote the center of the circle and r denote the radius.

Compare, (h,k) and (0,0) also radius ris 65 .

Here, h=0,k=0 and r=65 .

Substitute the values in standard equation of circle,

(x0)2+(y0)2=(65)2x2+y2=65

Thus, the equation of circle is x2+y2=65 .

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