Question

If a circle of constant radius 3k passes through the origin and meets the axes at $A$ and $B$, the locus of the centroid of $\Delta OAB$ is
A) ${x^2} + {y^2} = {k^2}$
B) ${x^2} + {y^2} = 2{k^2}$
C) ${x^2} + {y^2} = 3{k^2}$
D) ${x^2} + {y^2} = 4{k^2}$

Answer 1

Here we will first consider the general equation of the circle as the equation of the circle of the given constant radius. Then we will use the formula of radius to find the value of the variables. From there, we will find the locus of the centroid of the required triangle.

Complete step by step solution:
Let the equation of the circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$ of constant radius $3k$ passing through the origin.
Since the circle is passing through the origin , therefore $\left( {x,y} \right) = \left( {0,0} \right)$ .
We will put the value of the coordinate of origin, $\left( {x,y} \right) = \left( {0,0} \right)$, in the equation circle. Therefore, we get
$\begin{array}{l} \Rightarrow {0^2} + {0^2} + 2 \times g \times 0 + 2 \times f \times 0 + c = 0\\\ \Rightarrow c = 0\end{array}$
It is given that the radius of the given circle is $3k$.
We know the formula of radius of circle is given by
$\Rightarrow r = \sqrt {{g^2} + {f^2} - c}$
Substituting the value of radius, we get
$\Rightarrow 3k = \sqrt {{g^2} + {f^2} - c}$
Squaring both the sides, we get
$\begin{array}{l} \Rightarrow {\left( {3k} \right)^2} = {\left( {\sqrt {{g^2} + {f^2} - c} } \right)^2}\\\ \Rightarrow 9{k^2} = {g^2} + {f^2} - c\end{array}$
We know that the value of $c$ is zero.
Therefore,
$\Rightarrow 9{k^2} = {g^2} + {f^2} - 0$
$\Rightarrow 9{k^2} = {g^2} + {f^2}$ …… $\left( 1 \right)$
According to the question, the circle cut the coordinates at $A$ and $B$.
We will first find the point of intersection of the circle in the $x$-axis. For that, we will value $y$ as zero in the equation of the circle.
$\Rightarrow {x^2} + {0^2} + 2gx + 2 \times f \times 0 + c = 0$
We know that the value of $c$ is zero.
$\Rightarrow {x^2} + {0^2} + 2gx + 2 \times f \times 0 + 0 = 0$
On further simplification, we get
$\Rightarrow {x^2} + 2gx = 0$
On factoring the equation, we get
$\Rightarrow x\left( {x + 2g} \right) = 0$
This is possible only when $x + 2g = 0$ because $x$ can’t be zero.
Therefore,
$\Rightarrow x = - 2g$
Therefore, the coordinate of point $A$ is $\left( { - 2g,0} \right)$ .
Now, we will first find the point of intersection of the circle in the $y$-axis. For that, we will put the value of $x$ as zero in the equation of the circle.
$\Rightarrow {0^2} + {y^2} + 2 \times g \times 0 + 2fy + c = 0$
We know that the value of $c$ is zero.
$\Rightarrow {0^2} + {y^2} + 2 \times g \times 0 + 2fy + 0 = 0$
On further simplification, we get
$\Rightarrow {y^2} + 2fx = 0$
On factoring the equation, we get
$\Rightarrow y\left( {y + 2f} \right) = 0$
This is possible when $y + 2f = 0$ but $y$ can’t be zero.
Therefore,
$\Rightarrow y = - 2f$
Therefore, the coordinate of point $B$ is $\left( {0, - 2f} \right)$ .
Therefore, we have got the value of points:-
$O = \left( {0,0} \right)$
$A = \left( { - 2g,0} \right)$
$B = \left( {0, - 2f} \right)$
Let the coordinate of the centroid of the circle be $\left( {h',k'} \right)$
We know that the coordinates will be:-
$\begin{array}{l}h' = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}\\\k' = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}\end{array}$
Substituting the value of x coordinate of all three point in formula, we get
$\begin{array}{l} \Rightarrow h' = \dfrac{{ - 2g + 0 + 0}}{3} = \dfrac{{ - 2g}}{3}\\\ \Rightarrow k' = \dfrac{{ - 2f + 0 + 0}}{3} = \dfrac{{ - 2f}}{3}\end{array}$
Now, we will find the value of $f$ from here.
$\begin{array}{l} \Rightarrow h' = \dfrac{{ - 2g}}{3}\\\ \Rightarrow g = - \dfrac{{3h'}}{2}\end{array}$
Now, we will find the value of $g$ from here.
$\begin{array}{l} \Rightarrow k' = \dfrac{{ - 2f}}{3}\\\ \Rightarrow f = - \dfrac{{3k'}}{2}\end{array}$
Now, we will substitute the value of $f$ and $g$ in equation
$\Rightarrow 9{k^2} = {\left( { - \dfrac{{3h'}}{2}} \right)^2} + {\left( { - \dfrac{{3k'}}{2}} \right)^2}$
On finding the squares of the terms, we get
$\Rightarrow 9{k^2} = {\dfrac{{9h'}}{4}^2} + {\dfrac{{9k'}}{4}^2}$
On further simplification, we get
$\Rightarrow 4{k^2} = h{'^2} + k{'^2}$
Now, to find the locus of the centroid of the triangle, we will substitute the value of coordinate to be $x$ and $y$.
Thus, the locus of locus of the centroid of the $\Delta OAB$ is
$\begin{array}{l} \Rightarrow 4{k^2} = {x^2} + {y^2}\\\ \Rightarrow {x^2} + {y^2} = 4{k^2}\end{array}$

Hence, the correct option is option D.

Note:
Since here we have found the locus of the centroid of the triangle. A triangle is a two dimensional geometric shape which has 3 sides. Centroid is defined as a point which is formed by the intersection of all the medians of the triangle. Locus is defined as a way to represent the set of points in an equation.