Question

Find the condition that the straight line $cx-by+b^2=0$ may touch the circle $x^2+y^2=ax+by$ and find the point of contact.

Answer 1

The equation of the circle is $x^2+y^2-ax-by=0$
The line having equation $cx-by+b^2=0$ is a tangent to the circle.
Substituting the above equation into the circle, we have
$x^2+{\left(\frac{cx+b^2}{b}\right)}^2-ax-cx-b^2=0$
$\Rightarrow b^2{x}^2+c^2{x}^2+b^4+2cx{b}^2-ax{b}^2-cx{b}^2-b^4=0$
$\Rightarrow x^2(b^2+c^2)+x(c{b}^2-a{b}^2)=0$
For the line to be a tangent, the above quadratic equation must have only one solution.
$\therefore c{b}^2-a{b}^2=0$
Since $b\ne 0$, the required condition becomes $a=c$
$\therefore x^2(b^2+c^2)=0$ or $x=0$
$\Rightarrow 0-by+b^2=0$ or $y=b$
The point of contact is thus $(0,b)$