Key Concepts and Formulas

• General Equation of a Circle: The equation of a circle with center $(-g, -f)$ and radius $r$ is given by: $x^2 + y^2 + 2gx + 2fy + c = 0$
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
• Condition for Tangency: If two circles with centers $C_1$ and $C_2$ and radii $r_1$ and $r_2$ touch each other, then the distance between their centers is equal to the sum or difference of their radii: $|C_1C_2| = |r_1 \pm r_2|$. The plus sign is for external tangency, and the minus sign is for internal tangency.

Step-by-Step Solution

Step 1: Define the equation of the circle passing through (0,0) and (1,0).

Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since it passes through (0, 0), we have: $(0)^2 + (0)^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0$ So, the equation becomes $x^2 + y^2 + 2gx + 2fy = 0$.

Since it also passes through (1, 0), we have: $(1)^2 + (0)^2 + 2g(1) + 2f(0) = 0 \implies 1 + 2g = 0 \implies g = -\frac{1}{2}$ Thus, the equation of the circle is $x^2 + y^2 - x + 2fy = 0$. The center of this circle is $(\frac{1}{2}, -f)$ and its radius is $r = \sqrt{(\frac{1}{2})^2 + f^2} = \sqrt{\frac{1}{4} + f^2}$.

Step 2: Apply the condition for tangency with the given circle.

The given circle is $x^2 + y^2 = 9$, which has center (0, 0) and radius 3. Since the circle we found touches this circle, the distance between their centers must be equal to the sum or the difference of their radii. Thus, $\sqrt{\left(\frac{1}{2} - 0\right)^2 + (-f - 0)^2} = \left|3 \pm \sqrt{\frac{1}{4} + f^2}\right|$ $\sqrt{\frac{1}{4} + f^2} = \left|3 \pm \sqrt{\frac{1}{4} + f^2}\right|$

Step 3: Solve for f.

We have two cases to consider:

Case 1: $\sqrt{\frac{1}{4} + f^2} = 3 + \sqrt{\frac{1}{4} + f^2}$ This implies $3 = 0$, which is impossible.

Case 2: $\sqrt{\frac{1}{4} + f^2} = \left|3 - \sqrt{\frac{1}{4} + f^2}\right|$ Squaring both sides: $\frac{1}{4} + f^2 = 9 - 6\sqrt{\frac{1}{4} + f^2} + \frac{1}{4} + f^2$ $0 = 9 - 6\sqrt{\frac{1}{4} + f^2}$ $6\sqrt{\frac{1}{4} + f^2} = 9$ $\sqrt{\frac{1}{4} + f^2} = \frac{9}{6} = \frac{3}{2}$ Squaring both sides again: $\frac{1}{4} + f^2 = \frac{9}{4}$ $f^2 = \frac{9}{4} - \frac{1}{4} = \frac{8}{4} = 2$ $f = \pm\sqrt{2}$

Step 4: Determine the center of the circle.

The center of the circle is $(\frac{1}{2}, -f)$. Therefore, the possible centers are $(\frac{1}{2}, -\sqrt{2})$ and $(\frac{1}{2}, \sqrt{2})$. However, only $(\frac{1}{2}, -\sqrt{2})$ is among the options.

Step 5: Re-examine the logic. We made a mistake in Step 3, Case 2 when taking the absolute value. We need to consider when $3 - \sqrt{\frac{1}{4} + f^2}$ can be negative. In this case, $3 - \sqrt{\frac{1}{4} + f^2} = -\sqrt{\frac{1}{4} + f^2}$, so $3 = 0$, which is still impossible.

Let's re-examine the distance condition. The distance between the centers (1/2, -f) and (0, 0) is $\sqrt{1/4 + f^2}$. The radii are $\sqrt{1/4 + f^2}$ and 3. The tangency condition is $\sqrt{1/4 + f^2} = |3 \pm \sqrt{1/4 + f^2}|$. If we choose the negative sign, $\sqrt{1/4 + f^2} = |3 - \sqrt{1/4 + f^2}|$. We get $f = \pm \sqrt{2}$. However, the options do not contain $(\frac{1}{2}, \sqrt{2})$.

The question states that the circle touches the circle $x^2 + y^2 = 9$. This allows for either internal or external tangency. Let us assume the circle with center (1/2, -f) is inside the circle $x^2 + y^2 = 9$. Then we would have $|3 - \sqrt{1/4 + f^2}| = \sqrt{1/4 + f^2}$, which gives $f = \pm \sqrt{2}$.

If the circle is outside the circle $x^2 + y^2 = 9$, then $3 + \sqrt{1/4 + f^2} = \sqrt{1/4 + f^2}$, which is not possible.

The problem statement says the answer is $\left( \frac{1}{2}, \frac{1}{2} \right)$. So let's assume the center is $(\frac{1}{2}, \frac{1}{2})$. Then the radius is $\sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}}$. The distance between centers is $\sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}}$. Then $\sqrt{\frac{1}{2}} = |3 \pm \sqrt{\frac{1}{2}}|$. Then $\sqrt{\frac{1}{2}} = 3 - \sqrt{\frac{1}{2}}$. So $2\sqrt{\frac{1}{2}} = 3$, i.e. $\sqrt{2} = 3$, which is wrong.

We are told the answer is $(\frac{1}{2}, \frac{1}{2})$. If the center is $(\frac{1}{2}, \frac{1}{2})$, then $f = -\frac{1}{2}$, so $f^2 = \frac{1}{4}$. The radius is $\sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{2}}{2}$. The distance between (0, 0) and $(\frac{1}{2}, \frac{1}{2})$ is $\sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{2}}{2}$. So we want $\frac{\sqrt{2}}{2} = |3 - \frac{\sqrt{2}}{2}|$. This yields $\sqrt{2} = 3$, which is wrong.

There must be an error in the question or the answer. The correct approach is as above.

Common Mistakes & Tips

• Remember to consider both internal and external tangency when dealing with tangent circles.
• Be careful when squaring equations involving square roots; ensure you consider both positive and negative roots.
• Always check your solutions by plugging them back into the original equations.

Summary

We set up the general equation of a circle passing through (0, 0) and (1, 0). Then we applied the condition that this circle touches the circle $x^2 + y^2 = 9$. We obtained an equation involving the y-coordinate of the center. After simplifying and solving, we found the center of the circle to be $(\frac{1}{2}, -\sqrt{2})$. However, the given answer is $(\frac{1}{2}, \frac{1}{2})$, which is incorrect.

Final Answer

The final answer is $(\frac{1}{2}, -\sqrt{2})$. The provided answer $(\frac{1}{2}, \frac{1}{2})$ is incorrect. The closest option is (B), but even that is incorrect. There appears to be an error in the options provided.