Question 7

The condition $|V_1| = |V_2|$ implies: $\frac{k|q|}{r_1} = \frac{k|mq|}{r_2} \implies r_2 = |m|r_1$

Let the coordinates of the point be $(x, y)$. Then: $(x-ma)^2 + (y-mb)^2 = m^2 \left[ (x-a)^2 + (y-b)^2 \right]$

Expanding both sides: $x^2 - 2max + m^2a^2 + y^2 - 2mby + m^2b^2 = m^2(x^2 - 2ax + a^2 + y^2 - 2by + b^2)$ $x^2 + y^2 - 2max - 2mby = m^2x^2 + m^2y^2 - 2m^2ax - 2m^2by$

Rearranging the terms to one side: $(m^2-1)x^2 + (m^2-1)y^2 - 2ax(m^2-m) - 2by(m^2-m) = 0$

Since $m \neq 1$, we can divide the entire equation by $(m-1)$: $(m+1)x^2 + (m+1)y^2 - 2amx - 2bmy = 0$

Now let's check the given options: (A) For $m=-1$: $0x^2 + 0y^2 - 2a(-1)x - 2b(-1)y = 0 \implies 2ax + 2by = 0 \implies ax + by = 0$. (A straight line). Correct.

(B) For $m=2$: $3x^2 + 3y^2 - 4ax - 4by = 0 \implies x^2 + y^2 - \frac{4}{3}ax - \frac{4}{3}by = 0$. This is a circle with Center $= (\frac{2a}{3}, \frac{2b}{3})$ and Radius $R = \sqrt{\left(\frac{2a}{3}\right)^2 + \left(\frac{2b}{3}\right)^2} = \frac{2}{3}\sqrt{a^2+b^2}$. Correct.

(C) For $m=-2$: $-x^2 - y^2 + 4ax + 4by = 0 \implies x^2 + y^2 - 4ax - 4by = 0$. This is a circle with Center $= (2a, 2b)$ and Radius $R = \sqrt{(2a)^2 + (2b)^2} = 2\sqrt{a^2+b^2}$. Correct.

(D) For $m=-3$: The equation will retain $x^2$ and $y^2$ terms, representing a circle, not a straight line locus like $3bx+3ay=0$. Incorrect.

Correct Options: (A), (B), (C)