Question

Find the value of $p$ so that the straight line $x\cos \alpha +y\sin \alpha -p=-0$ may touch the circle $x^2+y^2-2ax\cos \alpha -2by\sin \alpha -a^2{\sin }^2\alpha =0$.

Answer 1

If a line touches a circle it means that the line is the tangent to the circle

So for a line to be tangent to a circle, the perpendicular distance of the centre of the circle from the line should be equal to the radius of the circle.

The center of the circle is $(a\cos a,b\sin a)$ and the radius is $\sqrt{a^2+b^2{\sin }^{2}a}$

So, from the question, we have

$\sqrt{a^2+b^2{\sin }^{2}a}=\frac{|a{\cos }^{2}a+b{\sin }^{2}a-p|}{\sqrt{{\cos }^{2}a+{\sin }^{2}a}}$

$p=\sqrt{a^2+b^2{\sin }^{2}a}+a{\cos }^{2}a+b{\sin }^{2}a$ or $p=a{\cos }^{2}a+b{\sin }^{2}a-\sqrt{a^2+b^2{\sin }^{2}a}$