Question

If a circle of constant radius 3k passes through the origin 0 and meets the coordinate axes at A and B, then the locus of the centroid of triangle OAB is
1. ${x^2} + {y^2} = {(2k)^2}$
2. ${x^2} + {y^2} = {(3k)^2}$
3. ${x^2} + {y^2} = {(4k)^2}$
4. ${x^2} + {y^2} = {(6k)^2}$

Answer 1

Hint: Consider the general equation of circle i.e.${x^2} + {y^2} + 2gx + 2fy + c = 0$ as the equation of the circle of constant radius 3k and also use the formula of radius of circle i.e. $r = \sqrt {{g^2} + {f^2} - c} $and find the value of x and y to find out the locus of the centroid of triangle OAB.

Complete step-by-step answer:
Let assume ${x^2} + {y^2} + 2gx + 2fy + c = 0$be the equation of a circle of constant radius 3k passing through the origin
Since circle is passing through the origin i.e. (x, y) = (0, 0)
Putting the value of x and y in equation of circle
\[ \Rightarrow \]c = 0
Since the radius of circle is constant
By the formula of radius of a circle $r = \sqrt {{g^2} + {f^2} - c} $
Since c = 0
Therefore $r = \sqrt {{g^2} + {f^2}} $
Putting the value of r and then squaring both sides
\[ \Rightarrow \]$9{k^2} = {g^2} + {f^2}$ (Equation 1)
According to the question circles cuts a coordinate axes at A and B
Let’s first check the intersection point of circle in the X-axes i.e. (x, 0)
\[ \Rightarrow \]${x^2} + 0 + 2gx + 2f \times 0 + c = 0$
\[ \Rightarrow \]${x^2} + 2gx = 0$
\[ \Rightarrow \]$x(x + 2g) = 0$
Since x can’t be 0
Therefore x = - 2g
Let assume that the A is the intersection point in the X-axes
Therefore A = (- 2g, 0)
Let’s check the intersection point of circle in the Y-axes i.e. (0, y)
\[ \Rightarrow \]$0 + {y^2} + 2g \times 0 + 2fy + c = 0$
\[ \Rightarrow \]${y^2} + 2fy = 0$
\[ \Rightarrow \]$y(y + 2f) = 0$
Since y can’t be 0
Therefore y = -2f
Let assume that the B is the intersection point in the Y-axes
Therefore B = (0, -2f)
A = (- 2g, 0)
B = (0, -2f)
O = (0, 0)
Let centroid of triangle OAB is G= $(h',k')$
Since we know that for x = $\dfrac{{{x_1} + {x_2} + {x_3}}}{3}$
 \[h' = \dfrac{{ - 2g + 0 + 0}}{3}\]
$h' = \dfrac{{ - 2g}}{3}$
$g = \dfrac{{ - 3h'}}{2}$
And for y =\[\dfrac{{{y_1} + {y_2} + {y_3}}}{3}\]
\[k' = \dfrac{{0 - 2f + 0}}{3}\]
$k' = \dfrac{{ - 2f}}{3}$
$f = \dfrac{{ - 3k'}}{2}$
Putting the values of f and g in equation 1
\[ \Rightarrow \]$9{k^2} = {(\dfrac{{ - 3h'}}{2})^2} + {(\dfrac{{ - 3k'}}{2})^2}$
\[ \Rightarrow \]$9{k^2} = \dfrac{{9{{h'}^2}}}{2} + \dfrac{{9{{k'}^2}}}{2}$
\[ \Rightarrow \]$4{k^2} = {h'^2} + {k'^2}$
Since we assumed the coordinates of centroid of triangle now we got a relation between them
Therefore locus of the centroid of triangle OAB is $4{k^2} = {x^2} + {y^2}$

Note: In these types of questions taking a general equation of circle since the circle passing through therefore c = 0 then using the formula of radius i.e. ($r = \sqrt {{g^2} + {f^2} - c} $) to find the equation and then finding the intersection point in the X and Y axes next assuming that A and B are the intersection point in the X axes and Y axes now assuming the centroid of triangle OAB are $(h',k')$now using the formula of x and y coordinates i.e. ($x = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}$ and \[y = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}\])to find the value of $(h',k')$since we got a relation between $(h',k')$and (x, y) therefore substituting the x and y by $(h',k')$hence we got the locus of the centroid of triangle OAB i.e. $4{k^2} = {x^2} + {y^2}$