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Question

To find the equation of tangent and normal to the circle x2+y23x+4y31=0 at the point (2,3).

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Solution

x2+y23x+4y31=0differentiatew.rtox2x+2ydy/dx3+4dy/dx=0dy/dx(2y+4)+2x3=0dy/dx=32x(2y+4)(dy/dx)2/3=346+4=110dxdy=10equationoftangent=(y3)=dy/dx(x2)y3=110(x2),10y30=x+2,x+10y=32equationofnormal(y3)=dxdy(x2),y3=+10(x2)10xy=17

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