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Question

Find the equation of the circle which passes through the point (2,1) and is tangent to the line 3x2y6=0 at the point (4,3).

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Solution

3x2y6=0

32dydx=0

dydx=32

x2x0y2y0=32

x0=3x213x22+3(y1y2)2+4x2(y2y1)6(x1x2)4y1+4y2=27

y0=(x1x2)2+y21+3(x1+x2)y2y223(x2x1)+2(y1y2)=417

r2=(x1x2)2+(y1y0)2

r=10137

(x+27)2+(y417)2=130049

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