Question

Show that the straight line $7y-x=5$ touches the circle $x^2+y^2-5x+5y=0$ and find the co-ordinates of the point of contact. Show that the other parallel tangent is $7y-x=-45$.

Answer 1

Circle is $(\frac{5}{2},\frac{-5}{2}),\frac{5}{2}\sqrt{2}$
Line is $7y-x-5=0$.
Condition of tangency $p=r$
$p=\frac{-50}{2\sqrt{\left(50\right)}}=\frac{5\sqrt{2}}{2}=r$
Parallel tangent : Any line parallel to
$7y-x-5=0$ is $7y-x+\lambda =0$
Condition of tangency
$\frac{7(-5/2)-5/2+\lambda }{\sqrt{\left(50\right)}}=\frac{5\sqrt{2}}{2}$
$\lambda -20=\pm 5\sqrt{2}.5\sqrt{2}/2=\pm 25$.
$\therefore \lambda =-5$ or $+45$.
Hence the other parallel tangent is
$7y-x+45=0$ as $7y-x-5=0$
is already a tangent.
Point of contact $P$ is foot of perpendicular from centre $C$ on the tangent at $P$. If its co-ordinates be $(h,k)$, then
$\frac{h-5/2}{-1}=\frac{k+5/2}{7}=-\frac{t}{a^2+b^2}(cor.)$
$=-\frac{(-5/2)+7.(-5/2)-5}{1^2+7^2}=\frac{25}{50}=\frac{1}{2}$
$\therefore h=\frac{5}{2}-\frac{1}{2},k=-\frac{5}{2}+\frac{7}{2}=1$
$\therefore P$ is $(2,1)$.
Follow this method for point of contact.
$2nd$ Method : Solve the tangent with the circle by eliminating one variable say $y$ and you will have a quadratic in $x$ having equal roots as the line is a tangent giving $x=2$ and hence $y=1$.
$\therefore$ Point is $(2,1)$.