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 line 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 circle is $x^2+y^2=4$
Assuming $P=(2\cos \theta ,2\sin \theta )$


Let the midpoint of $MN=(h,k)$
Tangent at $P$,
$2\cos \theta \cdot x+2\sin \theta \cdot y-4=0\Rightarrow x\cos \theta +y\sin \theta -2=0$
So, the coordinates of $M$ and $N$ are,
$M=(\frac{2}{\cos \theta },0)N=(0,\frac{2}{\sin \theta })$

Now, the midpoint is
$(h,k)=(\frac{1}{\cos \theta },\frac{1}{\sin \theta })\Rightarrow \frac{1}{h^2}+\frac{1}{k^2}=1$

Hence, the midpoint of $MN$ lies on the curve
$x^2+y^2=x^2{y}^2$