Point P moves on the circle $(x-6)^2 + y^2 = 9$. Find the locus of point Q which divides the line segment OP in the ratio 2:1.

1. Problem Description

Point P moves on the circle $(x-6)^2 + y^2 = 9$. Find the locus of point Q which divides the line segment OP in the ratio 2:

1.

2. Solution Steps

Let the coordinates of point P be $(X, Y)$, and the coordinates of point Q be $(x, y)$.

Since P moves on the circle $(x-6)^2 + y^2 = 9$, we have

$(X-6)^2 + Y^2 = 9$ (1)

Point Q divides the line segment OP in the ratio 2:

1. Then, the position vector of Q is given by:

$\vec{OQ} = \frac{2\vec{OP} + 1\vec{OO}}{2+1} = \frac{2\vec{OP}}{3}$

So,

$(x, y) = \frac{2}{3}(X, Y)$

$x = \frac{2}{3}X$ and $y = \frac{2}{3}Y$

Therefore,

$X = \frac{3}{2}x$ and $Y = \frac{3}{2}y$

Substitute $X = \frac{3}{2}x$ and $Y = \frac{3}{2}y$ into equation (1):

$(\frac{3}{2}x - 6)^2 + (\frac{3}{2}y)^2 = 9$

$(\frac{3x-12}{2})^2 + \frac{9}{4}y^2 = 9$

$\frac{9}{4}(x-4)^2 + \frac{9}{4}y^2 = 9$

Divide by $\frac{9}{4}$:

$(x-4)^2 + y^2 = 4$

3. Final Answer

The locus of point Q is $(x-4)^2 + y^2 = 4$.

This is a circle with center (4, 0) and radius 2.