Question

Let S be the circle in the xy-plane defined by the equation $x^2+y^2=4$. Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the segment MN must lie on the curve.

• A. $(x+y{)}^2=3xy$
• B. $x^{2/3}+y^{2/3}=2^{4/3}$
• C. $x^2+y^2=2xy$
• D. $x^2+y^2=x^2{y}^2$

Answer 1

The correct option is D $x^2+y^2=x^2{y}^2$

Given equation of circle is $x^2+y^2=4$

Equation of tangent to $S$ is given by

$\frac{x{x}_1}{4}+\frac{y{y}_1}{4}=1$ ... $\left(i\right)$ which is same as $\frac{x}{2h}+\frac{y}{2k}=1$ ... $\left(ii\right)$

Tangent at $P(2\cos \theta ,2\sin \theta )$ is $x\cos \theta +y\sin \theta =2$
$\Rightarrow M(2sec\theta ,0)$ and $N(0,2cosec\theta )$ .... from $\left(i\right)$ and $\left(ii\right)$
Midpoint is $(h,k)$
$\Rightarrow h=sec\theta ,k=cosec\theta$
$\Rightarrow \frac{1}{h^2}+\frac{1}{k^2}=1$

$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}=1$

$\Rightarrow x^2+y^2=x^2{y}^2$.