# The equation of a circle whose end points of diameter are P ( − 1 , 1 ) and Q ( 5 , 9 ) . ### 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 97E
To determine

## To calculate: The equation of a circle whose end points of diameter are P(−1,1) and Q(5,9) .

Expert Solution

The equation of the circle is (x2)2+(y5)2=25 .

### Explanation of Solution

Given information:

The end points of diameter of the circle are P(1,1) and Q(5,9) .

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(1,1) and Q(5,9) .

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(1,1) and Q(5,9) .

(x1+x22,y1+y22)=(1+52,1+92)=(42,102)=(2,5)

Therefore, coordinates of center are (2,5) .

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 (2,5) and (1,1) .

d=(12)2+(15)2=9+16=25=5

Therefore, radius of circle is 5.

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 (2,5) also radius r is 5.

Here, h=2,k=5 and r=5 .

Substitute the values in standard equation of circle,

(x2)2+(y5)2=52(x2)2+(y5)2=25

Thus, the equation of circle is (x2)2+(y5)2=25 .

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!