Key Concepts and Formulas

• Distance between centers of touching circles: If two circles touch internally, the distance between their centers is the absolute difference of their radii. If they touch externally, the distance is the sum of their radii.
• Equation of a circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
• Locus: The locus of a point is the path traced by the point as it moves according to given conditions.

Step-by-Step Solution

Step 1: Define the given circles and the variable circle.

Let $S_1$ be the circle $x^2 + y^2 = 9$, which has center $C_1 = (0, 0)$ and radius $r_1 = 3$. Let $S_2$ be the circle $(x - 2)^2 + y^2 = 1$, which has center $C_2 = (2, 0)$ and radius $r_2 = 1$. Let $S$ be the variable circle with center $P = (h, k)$ and radius $r$.

Step 2: Apply the conditions of tangency.

Since $S$ touches $S_1$ internally, the distance between their centers is the difference of their radii: $PC_1 = |r_1 - r| = r_1 - r$ (since $S$ is inside $S_1$, $r < r_1$). Therefore, $PC_1 = 3 - r$. $\sqrt{(h - 0)^2 + (k - 0)^2} = 3 - r$ $\sqrt{h^2 + k^2} = 3 - r$.

Since $S$ touches $S_2$ externally, the distance between their centers is the sum of their radii: $PC_2 = r + r_2 = r + 1$. $\sqrt{(h - 2)^2 + (k - 0)^2} = r + 1$. $\sqrt{(h - 2)^2 + k^2} = r + 1$.

Step 3: Eliminate the variable radius r.

From the first equation, $r = 3 - \sqrt{h^2 + k^2}$. Substitute this into the second equation: $\sqrt{(h - 2)^2 + k^2} = (3 - \sqrt{h^2 + k^2}) + 1$ $\sqrt{(h - 2)^2 + k^2} = 4 - \sqrt{h^2 + k^2}$ $\sqrt{(h - 2)^2 + k^2} + \sqrt{h^2 + k^2} = 4$

Step 4: Simplify the equation and find the locus.

$\sqrt{(h - 2)^2 + k^2} = 4 - \sqrt{h^2 + k^2}$ Square both sides: $(h - 2)^2 + k^2 = 16 - 8\sqrt{h^2 + k^2} + h^2 + k^2$ $h^2 - 4h + 4 + k^2 = 16 - 8\sqrt{h^2 + k^2} + h^2 + k^2$ $-4h + 4 = 16 - 8\sqrt{h^2 + k^2}$ $8\sqrt{h^2 + k^2} = 4h + 12$ $2\sqrt{h^2 + k^2} = h + 3$ Square both sides again: $4(h^2 + k^2) = (h + 3)^2$ $4h^2 + 4k^2 = h^2 + 6h + 9$ $3h^2 - 6h + 4k^2 = 9$ $3(h^2 - 2h) + 4k^2 = 9$ $3(h^2 - 2h + 1) + 4k^2 = 9 + 3$ $3(h - 1)^2 + 4k^2 = 12$ $\frac{(h - 1)^2}{4} + \frac{k^2}{3} = 1$

Replace $(h, k)$ with $(x, y)$ to get the locus: $\frac{(x - 1)^2}{4} + \frac{y^2}{3} = 1$ This is the equation of an ellipse centered at $(1, 0)$.

Step 5: Check the given points.

We need to find which of the given points satisfy the equation of the ellipse.

(A) $\left( \frac{1}{2}, \pm \frac{\sqrt{5}}{2} \right)$: $\frac{(\frac{1}{2} - 1)^2}{4} + \frac{(\pm \frac{\sqrt{5}}{2})^2}{3} = \frac{(\frac{-1}{2})^2}{4} + \frac{\frac{5}{4}}{3} = \frac{\frac{1}{4}}{4} + \frac{5}{12} = \frac{1}{16} + \frac{5}{12} = \frac{3 + 20}{48} = \frac{23}{48} \neq 1$

Let's recalculate. $\frac{(x - 1)^2}{4} + \frac{y^2}{3} = 1$ (A) $\left( \frac{1}{2}, \pm \frac{\sqrt{5}}{2} \right)$: $\frac{(\frac{1}{2} - 1)^2}{4} + \frac{(\pm \frac{\sqrt{5}}{2})^2}{3} = \frac{(\frac{-1}{2})^2}{4} + \frac{\frac{5}{4}}{3} = \frac{\frac{1}{4}}{4} + \frac{5}{12} = \frac{1}{16} + \frac{5}{12} = \frac{3 + 20}{48} = \frac{23}{48}$. This is NOT 1. There must be an error in the options. $\frac{(x - 1)^2}{4} + \frac{y^2}{3} = 1$ Let's recheck our work. $2\sqrt{h^2 + k^2} = h + 3$ $4(h^2 + k^2) = h^2 + 6h + 9$ $3h^2 - 6h + 4k^2 = 9$ $3(h^2 - 2h + 1) + 4k^2 = 12$ $3(h - 1)^2 + 4k^2 = 12$ $\frac{(h - 1)^2}{4} + \frac{k^2}{3} = 1$

The ellipse equation is correct. Let's test the points. (A) $(\frac{1}{2}, \pm \frac{\sqrt{5}}{2})$ $\frac{(\frac{1}{2} - 1)^2}{4} + \frac{(\frac{\sqrt{5}}{2})^2}{3} = \frac{(\frac{-1}{2})^2}{4} + \frac{\frac{5}{4}}{3} = \frac{1}{16} + \frac{5}{12} = \frac{3 + 20}{48} = \frac{23}{48} \neq 1$.

Typo: $3x^2+4y^2-6x-9=0$ If the point is $(\frac{1}{2}, \pm \frac{\sqrt{5}}{2})$, $3(\frac{1}{4}) + 4(\frac{5}{4}) - 6(\frac{1}{2}) - 9 = \frac{3}{4} + 5 - 3 - 9 = \frac{3}{4} - 7 = \frac{3 - 28}{4} = -\frac{25}{4} \neq 0$.

Let's check (B) (1, $\pm$ 2): $\frac{(1 - 1)^2}{4} + \frac{(\pm 2)^2}{3} = 0 + \frac{4}{3} \neq 1$.

Let's check (C) $(2, \pm \frac{3}{2})$: $\frac{(2 - 1)^2}{4} + \frac{(\pm \frac{3}{2})^2}{3} = \frac{1}{4} + \frac{\frac{9}{4}}{3} = \frac{1}{4} + \frac{3}{4} = 1$.

Therefore, the correct point is $(2, \pm \frac{3}{2})$. The question has the wrong answer.

Common Mistakes & Tips

• Be careful with signs when dealing with internal and external tangency. Internal tangency involves the difference of radii, while external tangency involves the sum.
• Squaring equations can introduce extraneous solutions. Always check your final solutions in the original equations.
• Double-check your algebraic manipulations to avoid errors.

Summary

We determined the locus of the center of the variable circle to be an ellipse. We found the equation of the ellipse to be $\frac{(x - 1)^2}{4} + \frac{y^2}{3} = 1$. Then, by substituting the given points into the equation of the ellipse, we found that the point $\left(2, \pm \frac{3}{2}\right)$ satisfies the equation. The correct answer provided was incorrect.

Final Answer

The final answer is $\left(2, \pm \frac{3}{2}\right)$, which corresponds to option (C).