A circle of constant radius 2r passes through the origin and meets the axes in 'P' and 'Q' Locus of the centroid of the `trianglePOQ` is :

Options

A. `x^(2) + y^(2) = r^(2)`
B. `9(x^(2) + y^(2)) = 16 r^(2)`
C. `2(x^(2) + y^(2)) = r^(2)`
D. `3(x^(2) + y^(2)) = 8r^(2)`

Answer

To find the locus of the centroid of triangle \( POQ \) formed by the points where a circle of radius \( 2r \) intersects the axes and passes through the origin, we can follow these steps:

### Step 1: Equation of the Circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Since the circle passes through the origin \((0,0)\), substituting these coordinates into the equation gives:
\[ 0 + 0 + 0 + 0 + c = 0 \implies c = 0 \]
Thus, the equation simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 2: Radius Condition
The radius \( r \) of the circle is given by:
\[ \sqrt{g^2 + f^2 - c} = 2r \]
Since \( c = 0 \), we have:
\[ \sqrt{g^2 + f^2} = 2r \]
Squaring both sides results in:
\[ g^2 + f^2 = 4r^2 \tag{1} \]

### Step 3: Points of Intersection with Axes
Let the circle intersect the x-axis at point \( P(a, 0) \) and the y-axis at point \( Q(0, b) \). The coordinates of these points can be derived from the circle's equation:
1. For point \( P(a, 0) \):
\[ a^2 + 0 + 2ga + 0 = 0 \implies a(a + 2g) = 0 \]
This gives \( a = 0 \) or \( a = -2g \).

2. For point \( Q(0, b) \):
\[ 0 + b^2 + 0 + 2fb = 0 \implies b(b + 2f) = 0 \]
This gives \( b = 0 \) or \( b = -2f \).

### Step 4: Centroid of Triangle \( POQ \)
The centroid \( G \) of triangle \( POQ \) with vertices \( P(a, 0) \), \( O(0, 0) \), and \( Q(0, b) \) is given by:
\[ G\left(\frac{a + 0 + 0}{3}, \frac{0 + 0 + b}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}\right) \]

### Step 5: Substitute Values of \( a \) and \( b \)
From the previous steps, we have:
- \( a = -2g \)
- \( b = -2f \)

Thus, the coordinates of the centroid become:
\[ G\left(\frac{-2g}{3}, \frac{-2f}{3}\right) \]

### Step 6: Locus of the Centroid
To find the locus of point \( G \), we can express \( g \) and \( f \) in terms of the coordinates of the centroid:
Let \( x = \frac{-2g}{3} \) and \( y = \frac{-2f}{3} \). Rearranging gives:
\[ g = -\frac{3x}{2}, \quad f = -\frac{3y}{2} \]
Substituting these into equation (1):
\[ \left(-\frac{3x}{2}\right)^2 + \left(-\frac{3y}{2}\right)^2 = 4r^2 \]
This simplifies to:
\[ \frac{9x^2}{4} + \frac{9y^2}{4} = 4r^2 \]
Multiplying through by 4 gives:
\[ 9x^2 + 9y^2 = 16r^2 \]
Dividing by 9 results in:
\[ x^2 + y^2 = \frac{16r^2}{9} \]
This represents a circle with center at the origin and radius \( \frac{4r}{3} \).

### Final Answer
The locus of the centroid \( G \) of triangle \( POQ \) is:
\[ x^2 + y^2 = \frac{16r^2}{9} \]