Let S be the circle in the X-Y 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 tangents to S at P Intersect the coordinate axes at the points M and N. then the midpoint of the segment MN must lie on the curve.
A. \[{{(x+y)}^{2}}=3xy\]
B. \[{{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{2}^{\dfrac{4}{3}}}\]
C. \[{{x}^{2}}+{{y}^{2}}=2xy\]
D. \[{{x}^{2}}+{{y}^{2}}={{x}^{2}}{{y}^{2}}\]
VerifiedComplete step-by-step solution:
Given a circle whose equation is $x^{2}+y^{2}=4 . \mathrm{P}$ is a point on the circle $\mathrm{S}$ with both coordinates being positive, tangents at $\mathrm{P}$ Intersect the coordinate axes at the points $\mathrm{M}$ and $\mathrm{N}$, then we have to find the curve on which midpoint of the segment MN lies, so as we know directly that equation of tangent on circle is
$\dfrac{x x_{1}}{4}+\dfrac{y y_{1}}{4}=1 \ldots . .(1)$
using (if equation of circle is $x^{2}+y^{2}=r^{2}$ then equation of tangent will be $\dfrac{x x_{1}}{r^{2}}+\dfrac{y y_{1}}{r^{2}}=1$ )
here $x_{1}, y_{1}$ are coordinates of point $\mathrm{P}$
We know that general coordinates of point on circle of radius $r$ is $(r \cos \theta, r \sin \theta)$
So as per given condition coordinates of $\mathrm{P}$ is $(2 \cos \theta, 2 \sin \theta)$
Putting it in equation (1) we get
$\dfrac{x(2 \cos \theta)}{4}+\dfrac{y(2 \sin \theta)}{4}=1$
Which on further solving equals to $x \cos \theta+y \sin \theta=2$
Now to find coordinates of $\mathrm{M}$ and $\mathrm{N}$ we can put the value of $\mathrm{y}$ and $\mathrm{x}$ equals to 0 respectively.
So, we got coordinates of $\mathrm{M}\left(\dfrac{2}{\cos \theta}, 0\right), \mathrm{N}\left(0, \dfrac{2}{\sin \theta}\right)$
Now, we are going to find the midpoint of MN.
We know that, if we have two points say $A_{1}\left(x_{1}, y_{1}\right) \& A_{2}\left(x_{2}, y_{2}\right)$ then the midpoint of these two points is equal to:
$\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)$
Then using this midpoint formula to find the midpoint of $\mathrm{M}\left(\dfrac{2}{\cos \theta}, 0\right), \mathrm{N}\left(0, \dfrac{2}{\sin \theta}\right)$ we get,
$\left(\dfrac{2}{\dfrac{\cos \theta}+0}{2}, \dfrac{0+\dfrac{2}{\sin \theta}}{2}\right) \\
=\left(\dfrac{2}{2 \cos \theta}, \dfrac{2}{2 \sin \theta}\right)$
In the above expression, 2 will be cancelled out from the numerator and denominator and we get,
$\left(\dfrac{1}{\cos \theta}, \dfrac{1}{\sin \theta}\right)$
Hence, we got the coordinates of the midpoint as $\left(\dfrac{1}{\cos \theta}, \dfrac{1}{\sin \theta}\right)$. Now, let us assume the x coordinate of this midpoint by $h$ and y coordinate by $k$. We know one property that is $\sin ^{2} \theta+\cos ^{2} \theta=1$
So, we can replace $\sin \theta, \cos \theta$ with $\left(\dfrac{1}{h}, \dfrac{1}{k}\right)$
So, it will become $\dfrac{1}{h^{2}}+\dfrac{1}{k^{2}}=1$, replace $\mathrm{h}, \mathrm{k}$ with $\mathrm{x}, \mathrm{y}$ for general
$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=1$ which on further solving equals to $x^{2}+y^{2}=x^{2} y^{2}$
Hence answer is $x^{2}+y^{2}=x^{2} y^{2}$ option (D).
Note: Most of the students have doubt that how we have calculated equation of tangent $\dfrac{x x_{1}}{r^{2}}+\dfrac{y y_{1}}{r^{2}}=1$, so for that we assume general point as $x_{1}, y_{1}$ now to find the slope we have to take $\dfrac{d y}{d x}$ at this point $2 x+2 y \dfrac{d y}{d x}=0$ on solving gives $\dfrac{d y}{d x}=\dfrac{-x}{y}$, on point $x_{1}, y_{1}$ equals to $\dfrac{d y}{d x}=\dfrac{-x_{1}}{y_{1}}$ Using formula $\left(y-y_{1}\right)=\dfrac{d y}{d x}\left(x-x_{1}\right),\left(y-y_{1}\right)=\dfrac{-x_{1}}{y_{1}}\left(x-x_{1}\right)$
Then on solving further and writing $x_{1}{ }^{2}+y_{1}{ }^{2}=r^{2}$ we get equation $\dfrac{x x_{1}}{r^{2}}+\dfrac{y y_{1}}{r^{2}}=1$
