Question

Find the equation of the circle passing through the points $(4,3)$ and $(3,2)$ and touching the line $3x-y-17=10$.

Answer 1

Chord through $A(4,3)$ and $B(3,2)$ is
$x-y-1=0$.
The required circle is
$\underset{CircleonABasdiameter}{(x-4)(x-3)+(y-3)(y-2)}+\underset{chordAB}{\lambda (x-y-1)}=0$
or $x^2+y^2-7x-5y+18+\lambda (x-y-1)=0.....\left(1\right)$
$x^2+y^2+(\lambda -7)x-(\lambda +5)y-(\lambda -18)=0$
centre is $(-\frac{\lambda -7}{2},\frac{\lambda +5}{2})$
and $r^2={\left(\frac{\lambda -7}{2}\right)}^2+{\left(\frac{\lambda +5}{2}\right)}^2+(\lambda -18)=\frac{{\lambda }^2+1}{2}$
If the line $3x-y-17=0$ is a tangent then $p=r$ gives
$\frac{-3\left(\frac{\lambda -7}{2}\right)-\frac{\lambda +5}{2}-17}{\sqrt{9+1}}=r$
or $(-2\lambda -9{)}^2=10r^2$
or $4{\lambda }^2+36\lambda +81=10\left[\frac{{\lambda }^2+1}{2}\right]$
or ${\lambda }^2-36\lambda -76=0$
or $(\lambda -38)(\lambda +2)=0$
or $\lambda =-2,38$
Putting the value of $\lambda$ in $\left(1\right)$ the required circles are
$x^2+y^2-9x-3y+20=0$
and $x^2+y^2+31x-43y-20=0$.