Find the equation of a circle whose center is at

Sarah-Louise Prince 2022-03-03 Answered
Find the equation of a circle whose center is at (4, 1) and tangent to the circle x2+y28x6y+9=0. Draw the circles.
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Expert Answer

zakrwinpfo
Answered 2022-03-04 Author has 4 answers

We have to find the equation of circle having a center (4,1) and tangent to the circle x2+y28x6y+9=0
The equation of the given circle is x2+y28x6y+9=0
x28x+1616+y26y+99+9=0
(x4)2+(y3)216=0
(x4)2+(y3)2=16(x4)2+(y3)2=42
center (4,3) and radius R=4
Now we have to circles having centers C1(4,1) and C2(4,3)
both circles touch each other,
then C1C2=(44)2+(31)2
C1C2=(0)2+(2)2
C1C2=4
C1C2=2
Since C1C2<R
so the point (4,1) inside the circle.
let radius of circle having center C1 is r.
then Rr=C1C2
4r=2
r=24
r=2
r=2
equation of circle having a center (4,1) and radius r=2 is
(x4)2+(y1)2=22
x28x+16+y22y+1=4
x2+y28x2y+17=4
x2+y28x2y=417
x2+y28x2y+13=0 (required equation of circle)
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