Key Concepts and Formulas

• Equation of a Circle: A circle with center $(h,k)$ and radius $r$ has the equation $(x-h)^2 + (y-k)^2 = r^2$. A circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ has the equation $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
• Equation of a Tangent: The equation of the tangent to a circle at a given point can be found using various methods, including using the derivative or properties of perpendicularity.
• Section Formula: If a point P divides the line segment joining points A and B in the ratio m:n, then the coordinates of P are given by $P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.

Step-by-Step Solution

Step 1: Find the equation of circle $C_1$.

Since $C_1$ passes through the origin and has a diameter of 4 along the positive x-axis, its center is at $(2, 0)$ and its radius is 2. Therefore, the equation of $C_1$ is $(x-2)^2 + y^2 = 2^2$, which simplifies to $x^2 - 4x + y^2 = 0$.

Step 2: Find the coordinates of point A.

Point A is the intersection of the circle $C_1$ and the line $y = 2x$. Substitute $y = 2x$ into the equation of $C_1$: $x^2 - 4x + (2x)^2 = 0$ $x^2 - 4x + 4x^2 = 0$ $5x^2 - 4x = 0$ $x(5x - 4) = 0$

The solutions are $x = 0$ and $x = \frac{4}{5}$. $x = 0$ corresponds to the origin O, so the x-coordinate of A is $\frac{4}{5}$. Then, $y = 2x = 2\left(\frac{4}{5}\right) = \frac{8}{5}$. Therefore, the coordinates of A are $\left(\frac{4}{5}, \frac{8}{5}\right)$.

Step 3: Find the equation of circle $C_2$.

$C_2$ has OA as a diameter, where O is $(0, 0)$ and A is $\left(\frac{4}{5}, \frac{8}{5}\right)$. The equation of $C_2$ is given by $(x - 0)\left(x - \frac{4}{5}\right) + (y - 0)\left(y - \frac{8}{5}\right) = 0$ $x^2 - \frac{4}{5}x + y^2 - \frac{8}{5}y = 0$ $5x^2 - 4x + 5y^2 - 8y = 0$

Step 4: Find the equation of the tangent to $C_2$ at point A.

The equation of $C_2$ can be written as $x^2+y^2 - \frac{4}{5}x - \frac{8}{5}y = 0$. The equation of the tangent at $A(x_1, y_1)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. Here, $x_1 = \frac{4}{5}$ and $y_1 = \frac{8}{5}$, $2g = -\frac{4}{5}$ and $2f = -\frac{8}{5}$. Therefore, the tangent at A is $x\left(\frac{4}{5}\right) + y\left(\frac{8}{5}\right) - \frac{2}{5}\left(x + \frac{4}{5}\right) - \frac{4}{5}\left(y + \frac{8}{5}\right) = 0$. $\frac{4}{5}x + \frac{8}{5}y - \frac{2}{5}x - \frac{8}{25} - \frac{4}{5}y - \frac{32}{25} = 0$ $\frac{2}{5}x + \frac{4}{5}y = \frac{40}{25} = \frac{8}{5}$ $2x + 4y = 8$ $x + 2y = 4$

Step 5: Find the coordinates of points P and Q.

Point P is the intersection of the tangent with the x-axis ($y = 0$). Substituting $y = 0$ into $x + 2y = 4$, we get $x = 4$. So, P is $(4, 0)$. Point Q is the intersection of the tangent with the y-axis ($x = 0$). Substituting $x = 0$ into $x + 2y = 4$, we get $2y = 4$, so $y = 2$. So, Q is $(0, 2)$.

Step 6: Calculate the ratio QA : AP.

$Q = (0, 2)$, $A = \left(\frac{4}{5}, \frac{8}{5}\right)$, and $P = (4, 0)$. Using the section formula, let A divide QP in the ratio $m:n$. Then $\frac{4}{5} = \frac{m(4) + n(0)}{m+n} \Rightarrow 4m = \frac{4}{5}(m+n) \Rightarrow 5m = m+n \Rightarrow 4m = n$. $\frac{8}{5} = \frac{m(0) + n(2)}{m+n} \Rightarrow 8m = \frac{2n}{m+n} \Rightarrow 8(m+n) = 10n \Rightarrow 4m+4n = 5n \Rightarrow 4m = n$. Thus $n = 4m$, and the ratio $m:n$ is $m:4m = 1:4$. Since A divides QP in the ratio 1:4, then QA : AP = 1 : 4.

Common Mistakes & Tips

• Be careful with signs when using the equation of a circle or tangent.
• Drawing a diagram can help visualize the problem and avoid errors.
• Remember the section formula correctly; it's a common source of mistakes.

Summary

We started by finding the equations of the two circles $C_1$ and $C_2$. We found the intersection point A of $C_1$ and the line $y=2x$. Then, we found the equation of the tangent to $C_2$ at point A. Next, we found the coordinates of points P and Q where the tangent intersects the x and y axes. Finally, we calculated the ratio QA : AP to be 1 : 4.

Final Answer

The final answer is \boxed{1 : 4}, which corresponds to option (A).