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

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)

$\frac{k+1}{h-2}=-\frac{3}{2}\phantom{\rule{0ex}{0ex}}3h+2k=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}equationofcircleis\phantom{\rule{0ex}{0ex}}(x-h{)}^{2}+(y-k{)}^{2}=(2-h{)}^{2}+(-1-k{)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}passesthrough(4,1),so\phantom{\rule{0ex}{0ex}}(4-h{)}^{2}+(1-k{)}^{2}=(2-h{)}^{2}+(-1-k{)}^{2}\phantom{\rule{0ex}{0ex}}-8h+16-2k+1=-4h+4+2k+1\phantom{\rule{0ex}{0ex}}4h-4k=12\phantom{\rule{0ex}{0ex}}h-k=3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}andfromabove\phantom{\rule{0ex}{0ex}}3h+2k=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}solvingwegeth=2/5,k=-13/5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}circleis\phantom{\rule{0ex}{0ex}}(x-h{)}^{2}+(y-k{)}^{2}=(2-h{)}^{2}+(-1-k{)}^{2}\phantom{\rule{0ex}{0ex}}(x-2/5{)}^{2}+(y+13/5{)}^{2}=(2-2/5{)}^{2}+(-1+13/5{)}^{2}=128/25\phantom{\rule{0ex}{0ex}}$

Regards

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