Key Concepts and Formulas

• General Equation of a Circle: $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$.
• Circle Touching the y-axis: If a circle touches the y-axis, then $c = f^2$.
• Midpoint Formula: The midpoint of the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Step-by-Step Solution

Step 1: Define the General Equation of the Circle

Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0 \quad \ldots(1)$ Why? This is the general form of a circle's equation. We will determine the parameters $g$, $f$, and $c$ based on the given conditions.

Step 2: Apply the Condition that the Circle Passes Through A(p, q)

Since the circle passes through $A(p, q)$, we substitute $x = p$ and $y = q$ into equation (1): $p^2 + q^2 + 2gp + 2fq + c = 0 \quad \ldots(2)$ Why? This equation enforces that the circle must pass through the point (p, q), establishing a relationship between $g$, $f$, $c$, $p$, and $q$.

Step 3: Apply the Condition for Touching the y-axis

Since the circle touches the y-axis, we have $c = f^2$. Substitute this into equation (2): $p^2 + q^2 + 2gp + 2fq + f^2 = 0 \quad \ldots(3)$ Why? This incorporates the tangency condition. By substituting $c = f^2$, we reduce the number of independent parameters.

Step 4: Relate the Center to the Endpoints of the Diameter

Let the other end of the diameter through $A(p, q)$ be $B(h, k)$. The center of the circle $(-g, -f)$ is the midpoint of $AB$. Therefore: $-g = \frac{p + h}{2} \quad \text{and} \quad -f = \frac{q + k}{2}$ Solving for $g$ and $f$: $g = -\frac{p + h}{2} \quad \ldots(4a)$ $f = -\frac{q + k}{2} \quad \ldots(4b)$ Why? We need to find the locus of $(h, k)$. Equations (4a) and (4b) express $g$ and $f$ in terms of $h$, $k$, $p$, and $q$, allowing us to substitute them into equation (3).

Step 5: Substitute and Simplify to Find the Locus

Substitute equations (4a) and (4b) into equation (3): $p^2 + q^2 + 2p\left(-\frac{p + h}{2}\right) + 2q\left(-\frac{q + k}{2}\right) + \left(-\frac{q + k}{2}\right)^2 = 0$ Simplify: $p^2 + q^2 - p(p + h) - q(q + k) + \frac{(q + k)^2}{4} = 0$ $p^2 + q^2 - p^2 - ph - q^2 - qk + \frac{q^2 + 2qk + k^2}{4} = 0$ $-ph - qk + \frac{q^2 + 2qk + k^2}{4} = 0$ Multiply by 4: $-4ph - 4qk + q^2 + 2qk + k^2 = 0$ $k^2 - 2qk + q^2 - 4ph = 0$ $(k - q)^2 = 4ph$ Why? This step eliminates $g$ and $f$ from the equation, leaving an equation in terms of $h$, $k$, $p$, and $q$. Further simplification leads to the locus.

Step 6: Express the Locus

Replace $(h, k)$ with $(x, y)$ to express the locus: $(y - q)^2 = 4px$ Why? This is the final step, where we replace the variables $(h, k)$ with the general coordinates $(x, y)$ to obtain the equation of the locus.

Common Mistakes & Tips

• Confusing the x-axis and y-axis tangency conditions. Remember $c=g^2$ for x-axis and $c=f^2$ for y-axis tangency.
• Careless algebraic manipulation. Double-check each step to avoid sign errors.
• Forgetting to replace $(h, k)$ with $(x, y)$ at the end to express the locus.

Summary

We started with the general equation of a circle and applied the given conditions: passing through a fixed point and touching the y-axis. We then used the midpoint formula to relate the center of the circle to the endpoints of a diameter. Substituting and simplifying, we obtained the locus of the other end of the diameter as $(y - q)^2 = 4px$.

Final Answer

The final answer is \boxed{(y - q)^2 = 4px}, which corresponds to option (A).