Question

A circle passes through $(-1,1),(0,6)$ and $(5,5)$. Find the points on this circle at which the tangents are parallel to the line joining the origin to its centre.

Answer 1

The required circle $x^2+y^2-4x-6y=0$ whose centre is $C(2,3)$. If $(h,k)$ be any point on this circle, then
$h^2+k^2-4h-6k=0$
Tangent at $(h,k)$ is
$x.h+y.k-2(x+h)-3(y+k)=0$
Its slope is $-\frac{h-2}{k-2}=\frac{3}{2}$ as it is parallel to join of $O$ and $C$.
$\therefore 2h+3k=13......\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get the points $(5,1)$ and $(-1,5)$
Alt. Method: $(x-2{)}^2+(y-3{)}^2=13$
or $X^2+Y^2=13$
Any tangent whose slope is $\frac{3}{2}$ i.e. of $OC$ is
$Y=mX\pm a\sqrt{1+m^2}$
or $y-3=\frac{3}{2}(x-2)\pm \sqrt{13}\sqrt{1+\frac{9}{4}}$
or $3x-2y+13=0$ or $3x-2y-13=0$
are two parallel tangents.
The points of contact are feet of perpendiculars from centre $(2,3)$ on these tangents. If it be $(h,k)$, then
$\frac{h-2}{3}=\frac{k-3}{-2}$
$=-\frac{3\left(2\right)-2\left(3\right)\pm 13}{3^2+2^2}=1$ or $-1$
$\therefore (h,k)$ is $(5,1)$ or $(-1,5)$.