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Chapter 11: Q.256 (page 1156)

In the following exercises, write the standard form of the equation of the circle with the given information:

Center is(3,6)and a point on the circle is(3,-2)

Short Answer

Expert verified

The equation of the circle isx2-6x+y2-12y=19

Step by step solution

01

Given information

The center is at(3,6)and a point on the circle is(3,-2)

02

We use the distance formula to calculate the radius of the circle

Let (x1,y1)=(3,6)(x2,y2)=(3,-2)

The distance formula can be given as

d=(x2-x1)2+(y2-y1)2

Substituting the values we get

d=(3-3)2+(-2-6)2d=0+(-8)2d=64d=卤8

As the radius cannot be negative

The radius is8units

03

use the standard form of equation and substitute the corresponding values 

The standard form of equation is

(x-h)2+(y-k)2=r2

Substituting the values we get

role="math" localid="1645419720707" (x-3)2+(y-6)2=82x2-6x+9+y2-12y+36=64x2-6x+y2-12y=19

04

Conclusion

The equation of the circle isx2-6x+y2-12y=19

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