BuyFind

Precalculus: Mathematics for Calcu...

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

Precalculus: Mathematics for Calcu...

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

Solutions

Chapter 1.8, Problem 96E
To determine

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

Expert Solution

Answer to Problem 96E

Theequation of the circle is (x+1)2+(y5)2=130 .

Explanation of Solution

Given information:

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

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 (1,5) and passes through the point (4,6) .

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

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)=(4(1))2+(65)2=(3)2+(11)2=9+121=130

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 (1,5) also radius ris 130 .

Here, h=1,k=5 and r=130 .

Substitute the values in standard equation of circle,

  (x(1))2+(y5)2=(130)2(x+1)2+(y5)2=130

Thus, the equation of circle is (x+1)2+(y5)2=130 .

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