Key Concepts and Formulas

• Family of Circles: The equation of a circle passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$, where $\lambda$ is a real number (and $\lambda \neq -1$ for it to represent a circle).
• General Equation of a Circle: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$.
• A point lying on a curve: If a point $(x_1, y_1)$ lies on a curve represented by the equation $f(x, y) = 0$, then $f(x_1, y_1) = 0$.

Step-by-Step Solution

Step 1: Formulating the Equation of the Family of Circles

We are given two circles: $S_1: x^2 + y^2 - 6x = 0$ $S_2: x^2 + y^2 - 4y = 0$

We want to find a circle that passes through the intersection of these two circles. According to the concept of the family of circles, the equation of any such circle can be written as: $S_1 + \lambda S_2 = 0$ where $\lambda$ is a parameter. Substituting the given circle equations: $(x^2 + y^2 - 6x) + \lambda(x^2 + y^2 - 4y) = 0$ This equation represents the family of circles passing through the intersection of $S_1$ and $S_2$.

Step 2: Finding the Center of the Family of Circles

Our goal is to rewrite the equation of the family of circles in the standard form to identify its center. Expanding the equation from Step 1, we have: $x^2 + y^2 - 6x + \lambda x^2 + \lambda y^2 - 4\lambda y = 0$ Grouping like terms: $(1 + \lambda)x^2 + (1 + \lambda)y^2 - 6x - 4\lambda y = 0$ To get the standard form, we divide by $(1 + \lambda)$, assuming $\lambda \neq -1$: $x^2 + y^2 - \frac{6}{1 + \lambda}x - \frac{4\lambda}{1 + \lambda}y = 0$ Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify: $2g = -\frac{6}{1 + \lambda} \implies g = -\frac{3}{1 + \lambda}$ $2f = -\frac{4\lambda}{1 + \lambda} \implies f = -\frac{2\lambda}{1 + \lambda}$ The center of the circle is $(-g, -f)$, so the center $C$ is: $C \equiv \left( \frac{3}{1 + \lambda}, \frac{2\lambda}{1 + \lambda} \right)$

Step 3: Using the Center Condition to Determine $\lambda$

We are given that the center of the circle lies on the line $2x - 3y + 12 = 0$. Therefore, the coordinates of the center must satisfy this equation. Substituting the coordinates of $C$ into the line equation: $2\left(\frac{3}{1 + \lambda}\right) - 3\left(\frac{2\lambda}{1 + \lambda}\right) + 12 = 0$ Multiplying by $(1 + \lambda)$ to clear the fractions: $6 - 6\lambda + 12(1 + \lambda) = 0$ $6 - 6\lambda + 12 + 12\lambda = 0$ $18 + 6\lambda = 0$ $6\lambda = -18$ $\lambda = -3$

Step 4: Finding the Equation of the Specific Circle

Now that we have found $\lambda = -3$, we can substitute it back into the equation of the family of circles to find the equation of the specific circle we are looking for: $(x^2 + y^2 - 6x) + \lambda(x^2 + y^2 - 4y) = 0$ Substituting $\lambda = -3$: $(x^2 + y^2 - 6x) - 3(x^2 + y^2 - 4y) = 0$ $x^2 + y^2 - 6x - 3x^2 - 3y^2 + 12y = 0$ $-2x^2 - 2y^2 - 6x + 12y = 0$ Dividing by $-2$: $x^2 + y^2 + 3x - 6y = 0$ This is the equation of the circle whose center lies on the given line.

Step 5: Checking the Options

We are given four points and we need to find which point lies on the circle $x^2 + y^2 + 3x - 6y = 0$. We substitute each point into the equation and check if the equation is satisfied.

(A) $(-3, 1)$: $(-3)^2 + (1)^2 + 3(-3) - 6(1) = 9 + 1 - 9 - 6 = -5 \neq 0$ (B) $(1, -3)$: $(1)^2 + (-3)^2 + 3(1) - 6(-3) = 1 + 9 + 3 + 18 = 31 \neq 0$ (C) $(-1, 3)$: $(-1)^2 + (3)^2 + 3(-1) - 6(3) = 1 + 9 - 3 - 18 = -11 \neq 0$ (D) $(-3, 6)$: $(-3)^2 + (6)^2 + 3(-3) - 6(6) = 9 + 36 - 9 - 36 = 0$

Since the point $(-3, 6)$ satisfies the equation of the circle, the circle passes through the point $(-3, 6)$.

There appears to be an error in the provided answer. The correct answer based on the given options is (D) (-3, 6), not (A) (-3, 1).

Common Mistakes & Tips

• Forgetting to divide by $(1+\lambda)$: When finding the center of the circle, make sure the coefficients of $x^2$ and $y^2$ are 1.
• Incorrectly calculating $\lambda$: Double-check your algebra when solving for $\lambda$.
• Sign Errors: Pay close attention to signs when substituting coordinates into the circle equation.

Summary

We found the equation of the family of circles passing through the intersection of two given circles. Then, using the condition that the center of the required circle lies on a given line, we found the value of $\lambda$. Substituting this value back into the family of circles equation, we obtained the equation of the specific circle. Finally, by substituting the given options into the circle's equation, we found that the circle passes through the point (-3, 6).

Final Answer

The final answer is \boxed{(-3, 6)}, which corresponds to option (D).