# 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 38E
To determine

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

Expert Solution

Equation of circle is x2+y2+2x10y104=0

### Explanation of Solution

Given information: Centre (1,5) and passes through (4,6)

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:

Center,(h,k)=(1,5)

Equation of circle is given as

(xh)2+(yk)2=r2

Substituting the values in above equation:

(x(1))2+(y5)2=r2(x+1)2+(y5)2=r2(1)

Also, given that circle passes through the point (4,6), thus the point must satisfy equation (1),

(4+1)2+(65)2=r2r2=(3)2+(11)2r2=9+121r2=130

Substituting the value of r2 in equation (1) to get the equation of circle

(x+1)2+(y5)2=130x2+1+2x+y2+2510y=130x2+y2+2x10y+(1+25130)=0x2+y2+2x10y104=0

Conclusion:

Hence, equation of circle is x2+y2+2x10y104=0

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