Find equation of circle tangent to 2x-3y-7 at (2, -1) & passes through(4, 1)

Dear student

Let the centre of the circle be (h,k)

The circle is tangent to 2x-3y-7 at (2, -1)
slope of normal at (2,-1) which passes through centre (h,k) is -3/2 (perpendicular to tangent with slope 2/3)

radius = distance between points (h,k) and (2,-1)

Circle also passes through(4, 1)

k+1h-2=-323h+2k=4equation of circle is(x-h)2+(y-k)2=(2-h)2+(-1-k)2passes through (4,1), so(4-h)2+(1-k)2=(2-h)2+(-1-k)2  -8h+16-2k+1=-4h+4+2k+14h-4k=12h-k=3and from above3h+2k=4solving we get h=2/5, k=-13/5circle is (x-h)2+(y-k)2=(2-h)2+(-1-k)2(x-2/5)2+(y+13/5)2=(2-2/5)2+(-1+13/5)2 =128/25





Regards

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