Question

Find the equation of the circle each of which has radius $5$ and has tangent as the line $3x-4y+5=0$ at $(1,2)$.

Answer 1

The tangent is line $3x-4y+5=0$ at $(1,2)$ of the circle

Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$

Hence the centre of circle $=(-g,-f)$

and radius of circle $=\sqrt{g^2+f^2-c}$

The distance from centre of circle to the line $3x-4y+5=0$ is $5$

By equation $3x-4y+5=0\Rightarrow y=\frac{3}{4}x+\frac{5}{4}$

slope of tangent$=\frac{3}{4}$

therefore slope of normal$=\frac{-4}{3}$

$\Rightarrow \frac{2-(-f)}{1-(-g)}=\frac{-4}{3}\Rightarrow 3f+4g=-10⟶\left(2\right)$

solving $\left(1\right)$ and $\left(2\right)$ we get,

$g=4andf=2then,r=\sqrt{{(-4)}^2+2^2-c}\Rightarrow 5=\sqrt{{(-4)}^2+2^2-c}\Rightarrow c=-5$

therefore equation of circle is given as;

$x^2+y^2-8x+4y-5=0$