BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter T, Problem 2BDT
To determine

To find: The equation of the circle that has center at (1,4) and passes through the point (3,2).

Expert Solution

Answer to Problem 2BDT

The equation of the circle is (x+1)2+(y4)2=52.

Explanation of Solution

Formula used:

The equation of the circle is (xh)2+(yk)2=r2, where the center of the circle is at the point (h,k), r is the radius of the circle.

Calculation:

Suppose, the equation of the circle is (xh)2+(yk)2=r2.

It is given that, the center of the circle is at the point (1,4).

Thus, substitute h=1 and k=4 in (xh)2+(yk)2=r2 to obtain the equation of the straight line,

(x+1)2+(y4)2=r2

It is given that, the line passes through the point (2,5).

Thus, the radius of the circle is the distance between the point (2,5) and the radius. That is,

r2=(3(1))2+(24)2=(4)2+(6)2=16+36=52

Thus, substitute r2=52 in (x+1)2+(y4)2=r2 to obtain the equation of the straight line,

(x+1)2+(y4)2=52

Thus, the equation of the circle is (x+1)2+(y4)2=52.

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