# To find the equation of circle that satisfies the given condition

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter B, Problem 37E
To determine

## To calculate: To find the equation of circle that satisfies the given condition

Expert Solution

Equation of circle is x2+y26x+2y15=0

### Explanation of Solution

Given information: Centre (3,1) and radius 5

Formula Used:

Equation of circle with center as (h,k) and radius r is given as

(xh)2+(yk)2=r2

Calculation:

Given the below details:

Equation of circle is given as

(xh)2+(yk)2=r2

Substituting the values in above equation:

(x3)2+(y(1))2=52(x3)2+(y+1)2=52x2+96x+y2+1+2y=25x2+y26x+2y+(9+125)=0x2+y26x+2y15=0

Conclusion:

Hence, equation of circle is x2+y26x+2y15=0

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