Find the equation of a circle whose center is at (4, 1) and tangent to the circle x2+y2−8x−6y+9=0. Draw the circles.

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zakrwinpfo · Accepted Answer

We have to find the equation of circle having a center (4,1) and tangent to the circle x2+y2−8x−6y+9=0The equation of the given circle is x2+y2−8x−6y+9=0⇒x2−8x+16−16+y2−6y+9−⧸9+⧸9=0⇒(x−4)2+(y−3)2−16=0⇒(x−4)2+(y−3)2=16⇒(x−4)2+(y−3)2=42center (4,3) and radius R=4Now we have to circles having centers C1(4,1) and C2(4,3)both circles touch each other,then C1C2=(4−4)2+(3−1)2C1C2=(0)2+(2)2C1C2=4C1C2=2Since C1C2&amp;lt;Rso the point (4,1) inside the circle.let radius of circle having center C1 is r.then R−r=C1C24−r=2−r=2−4−r=−2r=2equation of circle having a center (4,1) and radius r=2 is(x−4)2+(y−1)2=22x2−8x+16+y2−2y+1=4x2+y2−8x−2y+17=4x2+y2−8x−2y=4−17x2+y2−8x−2y+13=0 (required equation of circle)

Find the equation of a circle whose center is at

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