Question

Let $AB$ be a chord of circle ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ subtending a right angle at the centre. Then the locus of the centroid of $\Delta PAB$ as $P$ moves on circle –

(a) A parabola
(b) A circle
(c) An ellipse
(d) A pair of straight lines

Answer 1

Hint: This question is based on circle’s properties and concept of locus. First of all, we assume point $Q$ as $\left( h,k \right)$, where $Q$ is centroid of $\Delta PAB$, for which we have to find the locus of centroid.

By using following properties, we calculate locus of centroid according to the question –

(i) If two lines of slope ${{m}_{1}}$ and ${{m}_{2}}$ are perpendicular to each other, then

${{m}_{1}}\times {{m}_{2}}=-1$

(ii) If centroid of triangle $\Delta ABC$ where $A\left( {{x}_{1}},{{y}_{1}} \right)$, $B\left( {{x}_{2}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ is $G\left( x,y \right)$, then

$x=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}$ and $y=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}$

Here, we assume chord$AB$, where $A\left( r\cos {{\theta }_{1}},r\sin {{\theta }_{1}} \right)$ and $B\left( r\cos {{\theta }_{2}},r\sin {{\theta }_{2}} \right)$, but according to the question, point $P$ is variable, which moves in circle. We assume $P$ is$\left( {{x}_{1}},{{y}_{1}} \right)$. Now by using properties and equation of circle, we eliminate variable, \[{{x}_{1}}\] and\[{{y}_{1}}\], and get equation in terms of$\left( h,k \right)$, because centroid of $\Delta PAB$ is assumed to be$\left( h,k \right)$.

Complete step-by-step answer:

Now, let us get started with the solution.

We have the equation of circle as ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ … (i)

Therefore, we get that the centre is $\left( 0,0 \right)$ and radius $=r$

So, the circle can be represented as –

According to question, chord $AB$ subtends right angle on centre $O\left( 0,0 \right)$. Let us assume $A$ and $B$ are parametric points of a circle. So, $A$ is $\left( r\cos {{\theta }_{1}},r\sin {{\theta }_{1}} \right)$ and $B$ is $\left( r\cos {{\theta }_{2}},r\sin {{\theta }_{2}} \right)$.

Now, according to the question, $OA$ and $OB$ made $90{}^\circ $ to each other. So, the multiplication of slopes of $OA$ and $OB$ is $-1$.

We know that, slope of line joining two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ $=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$.

So, we can find the slope of line $OA$ joining $O\left( 0,0 \right)$ and $A\left( r\cos {{\theta }_{1}},r\sin {{\theta }_{1}} \right)$ as \[=\dfrac{\left( r\sin {{\theta }_{1}}-1 \right)}{\left( r\cos {{\theta }_{1}}-1 \right)}\]

$\Rightarrow {{m}_{OA}}=\tan {{\theta }_{1}}$

Now, we can find the slope of line $OB$ joining $O\left( 0,0 \right)$ and $B\left( r\cos {{\theta }_{2}},r\sin {{\theta }_{2}} \right)$ as $=\dfrac{\left( r\sin {{\theta }_{2}}-1 \right)}{\left( r\cos {{\theta }_{2}}-1 \right)}$

$\Rightarrow {{m}_{OB}}=\tan {{\theta }_{2}}$

According to the question, $OA$ and $OB$ are perpendicular. So, we have condition as

${{m}_{OA}}\times {{m}_{OB}}=-1$

$\Rightarrow \tan {{\theta }_{1}}\times \tan {{\theta }_{1}}=-1$

\[\Rightarrow \dfrac{\sin {{\theta }_{1}}\times \sin {{\theta }_{2}}}{\cos {{\theta }_{1}}\times \cos {{\theta }_{2}}}=-1\]

\[\Rightarrow \cos {{\theta }_{1}}.\cos {{\theta }_{2}}+\sin {{\theta }_{1}}.\sin {{\theta }_{2}}=0\] … (ii)

Now as we know that centroid of $\Delta ABC$, where $A\left( {{x}_{1}},{{y}_{1}} \right)$, $B\left( {{x}_{2}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ is $G\left( x,y \right)$, then

$x=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}$ and $y=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}$

Now let us take a point $P$ which moves on circle is $\left( {{x}_{1}},{{y}_{1}} \right)$ and centroid of $\Delta PAB$, where $P\left( {{x}_{1}},{{y}_{1}} \right)$, $A\left( r\cos {{\theta }_{1}},r\sin {{\theta }_{1}} \right)$ and $B\left( r\cos {{\theta }_{2}},r\sin {{\theta }_{2}} \right)$ is $G\left( h,k \right)$, then

$h=\dfrac{{{x}_{1}}+r\cos {{\theta }_{1}}+r\cos {{\theta }_{2}}}{3}$

$k=\dfrac{{{y}_{1}}+r\sin {{\theta }_{1}}+r\sin {{\theta }_{2}}}{3}$

So, we have $3h={{x}_{1}}+r\left( \cos {{\theta }_{1}}+\cos {{\theta }_{2}} \right)$

$\Rightarrow 3h-r\left( \cos {{\theta }_{1}}+\cos {{\theta }_{2}} \right)={{x}_{1}}$

And for k, we have $3k={{y}_{1}}+r\left( \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right)$

$\Rightarrow 3k-r\left( \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right)={{y}_{1}}$

Now, we know that point $P\left( {{x}_{1}},{{y}_{1}} \right)$ moves on circle, so it satisfies equation of circle:

${{x}^{2}}+{{y}^{2}}={{r}^{2}}$

$\Rightarrow {{\left\{ 3h-r\left( \cos {{\theta }_{1}}+\cos {{\theta }_{2}} \right) \right\}}^{2}}+{{\left\{ 3k-r\left( \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right) \right\}}^{2}}={{r}^{2}}$

$\Rightarrow {{\left\{ h-\dfrac{r}{3}\left( \cos {{\theta }_{1}}+\cos {{\theta }_{2}} \right) \right\}}^{2}}+{{\left\{ k-\dfrac{r}{3}\left( \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right) \right\}}^{2}}=\dfrac{{{r}^{2}}}{9}$

Now if we consider formula:

$\tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{\tan {{\theta }_{1}}-\tan {{\theta }_{2}}}{1+\tan {{\theta }_{1}}.\tan {{\theta }_{2}}}$

By equation (ii), $\tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\to \infty $

$\Rightarrow {{\theta }_{1}}-{{\theta }_{2}}=90{}^\circ $

$\Rightarrow {{\theta }_{1}}={{\theta }_{2}}+90{}^\circ $

$\Rightarrow {{\theta }_{2}}={{\theta }_{1}}-90{}^\circ $

So, \[{{\left\{ h-\dfrac{r}{3}\left( \cos {{\theta }_{1}}+\cos \left( {{\theta }_{1}}-90{}^\circ \right) \right) \right\}}^{2}}+{{\left\{ k-\dfrac{r}{3}\left( \sin {{\theta }_{1}}+\sin \left( {{\theta }_{1}}-90{}^\circ \right) \right) \right\}}^{2}}=\dfrac{{{r}^{2}}}{9}\]

\[\Rightarrow {{\left\{ h-\dfrac{r}{3}\left( \cos {{\theta }_{1}}+\sin {{\theta }_{1}} \right) \right\}}^{2}}+{{\left\{ k-\dfrac{r}{3}\left( \sin {{\theta }_{1}}-\cos {{\theta }_{1}} \right) \right\}}^{2}}=\dfrac{{{r}^{2}}}{9}\]

Replace $h\to x$ and $k\to y$,

\[\Rightarrow {{\left\{ x-\dfrac{r}{3}\left( \cos {{\theta }_{1}}+\sin {{\theta }_{1}} \right) \right\}}^{2}}+{{\left\{ y-\dfrac{r}{3}\left( \sin {{\theta }_{1}}-\cos {{\theta }_{1}} \right) \right\}}^{2}}=\dfrac{{{r}^{2}}}{9}\]

This is the equation of the circle. So, the locus of the centroid of $\Delta PAB$ is a circle.

So, the correct answer is “Option (b)”.

Note: In this question, point $P$ is moving in a circle and $AB$ are fixed. So we have to eliminate the variables ${{x}_{1}}$ and \[{{y}_{1}}\]. But sometimes $AB$ will not be fixed, point $P$ will be fixed. Then we will have to eliminate ${{\theta }_{1}}$ and ${{\theta }_{2}}$, and ${{x}_{1}}$ and ${{y}_{1}}$ will be in the equation. So, students should take care of which is fixed, and which one is variable.