# The equation of a circle whose end points of diameter are P ( − 1 , 3 ) and Q ( 7 , − 5 ) .

### 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 98E
To determine

## To calculate: The equation of a circle whose end points of diameter are P(−1,3) and Q(7,−5) .

Expert Solution

The equation of the circle is (x3)2+(y+1)2=32 .

### Explanation of Solution

Given information:

The end points of diameter of the circle are P(1,3) and Q(7,5) .

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 .

Midpoint of the segment that joins two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by (x1+x22,y1+y22) .

Calculation:

Consider the provided conditions that end points of diameter of the circle are P(1,3) and Q(7,5) .

Now, the center of the circle is the midpoint of the diameter.

Recall that the midpoint of the segment that joins two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by (x1+x22,y1+y22) .

Evaluate the midpoint of the segment that joins P(1,3) and Q(7,5) .

(x1+x22,y1+y22)=(1+72,352)=(62,22)=(3,1)

Therefore, coordinates of center are (3,1) .

Now, the radius of the circle is distance between the center of the circle and any one end point of diameter.

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 .

Evaluate the distance between (3,1) and (1,3) .

d=(13)2+(3+1)2=16+16=32

Therefore, radius of circle is 32 .

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 (3,1) also radius r is 32 .

Here, h=3,k=1 and r=32 .

Substitute the values in standard equation of circle,

(x3)2+(y(1))2=(32)2(x3)2+(y+1)2=32

Thus, the equation of circle is (x3)2+(y+1)2=32 .

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