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 3BDT
To determine

To find: The center and radius of the circle with equation x2+y26x+10y+9=0.

Expert Solution

Answer to Problem 3BDT

The center of the circle is at the point (3,5) and the radius of the circle is 5.

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:

It is given that the equation of the circle is x2+y26x+10y+9=0.

Simplify the above equation as follows,

x2+y26x+10y+9=0x26x+9+y2+10y+2525=0(x3)2+(y+5)2=25(x3)2+(y+5)2=52

Compare the equations (x3)2+(y+5)2=52 and (xh)2+(yk)2=r2.

It is clear that h=3, k=5, and r=5.

The center of the circle is at the point (h,k), that is (3,5).

The radius of the circle is r=5.

Thus, the center of the circle is at the point (3,5) and the radius of the circle is 5.

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