Question

To find the equation of tangent and normal to the circle $x^2+y^2-3x+4y-31=0$ at the point $(2,3)$.

Answer 1

$x^2+y^2-3x+4y-31=0differentiatew.rtox\Rightarrow 2x+2ydy/dx-3+4dy/dx=0\Rightarrow dy/dx\left(2y+4\right)+2x-3=0\therefore dy/dx=\frac{3-2x}{\left(2y+4\right)}\Rightarrow {\left(dy/dx\right)}_{2/3}=\frac{3-4}{6+4}=\frac{-1}{10}\frac{dx}{dy}=-10\therefore equationof\tan gent=\Rightarrow \left(y-3\right)=dy/dx\left(x-2\right)\Rightarrow y-3=\frac{-1}{10}\left(x-2\right),10y-30=x+2,x+10y=32equationofnormal\Rightarrow \left(y-3\right)=\frac{dx}{dy}\left(x-2\right),y-3=+10\left(x-2\right)\Rightarrow 10x-y=17$