Question

If a variable tangent of any circle $x^2+y^2=a^2$ intersects the axis at P and Q, then prove that the locus of middle point R on P and Q is: $4x^2{y}^2=a^2(x^2+y^2)$.

Answer 1

The equation of tangent $T$ to the circle at any point $(a\cos \theta ,a\sin \theta )$ on the circle is $x\cos \theta +y\sin \theta =a$.

Now, the points $P$ and $Q$ are intercepts of $T$ on x-axis and y-axis respectively.

The x-intercept (y=0) is $x\cos \theta +0\times \sin \theta =a⟹x\cos \theta =a⟹x=\frac{a}{\cos \theta }$

The y-intercept (x=0) is $y\sin \theta +0\times \cos \theta =a⟹y\sin \theta =a⟹y=\frac{a}{\sin \theta }$

$\therefore P=(\frac{a}{\cos \theta },0)$ and $Q=(0,\frac{a}{\sin \theta })$

Now, mid-point of $PQ$ is $R$ so, $R=(\frac{a}{2\cos \theta },\frac{a}{2\sin \theta })$

Now to find the locus of $R$, we need to eliminate the parameter $\theta$.

So, for any general point $(x,y)$ on locus of R we have

$\frac{1}{x^2}+\frac{1}{y^2}=\frac{4{\cos }^2\theta }{a^2}+\frac{4{\sin }^2\theta }{a^2}$ (squaring and adding inverse of coordinates)

$\therefore \frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{a^2}(\because {\sin }^{2}x+{\cos }^{2}x=1)$

$\therefore a^2(x^2+y^2)=4x^2{y}^2$

Hence proved.