A circle $$C$$ touches the line $$x = 2y$$ at the point $$(2, 1)$$ and intersects the circle $$C_1 : x^2 + y^2 + 2y - 5 = 0$$ at two points $$P$$ and $$Q$$ such that $$PQ$$ is a diameter of $$C_1$$. Then the diameter of $$C$$ is:

Circle $$C$$ touches the line $$L: x - 2y = 0$$ at the point $$(2, 1)$$. The equation of a family of circles touching a line $$L = 0$$ at a specific point $$(x_1, y_1)$$ is given by:

$$(x - x_1)^2 + (y - y_1)^2 + \lambda(L) = 0$$

Substituting our values:

$$(x - 2)^2 + (y - 1)^2 + \lambda(x - 2y) = 0$$

$$x^2 - 4x + 4 + y^2 - 2y + 1 + \lambda x - 2\lambda y = 0$$

$$x^2 + y^2 + x(\lambda - 4) - y(2\lambda + 2) + 5 = 0 \quad \text{--- (Equation of Circle } C)$$

The equation of the common chord between Circle $$C$$ and Circle $$C_1: x^2 + y^2 + 2y - 5 = 0$$ is found by subtracting their equations ($$C - C_1 = 0$$):

$$(x^2 + y^2 + x(\lambda - 4) - y(2\lambda + 2) + 5) - (x^2 + y^2 + 2y - 5) = 0$$

$$x(\lambda - 4) - y(2\lambda + 4) + 10 = 0 $$

We are told that $PQ$ is a diameter of circle $$C_1$$. This means the center of circle $$C_1$$ must lie on the line $$PQ$$.

• Center of $$C_1$$: From $$x^2 + y^2 + 2y - 5 = 0$$, the center is $$(0, -1)$$.

Substituting $$(0, -1)$$ into the equation for chord $$PQ$$:

$$0(\lambda - 4) - (-1)(2\lambda + 4) + 10 = 0$$

$$2\lambda + 4 + 10 = 0$$

$$2\lambda = -14 \implies \lambda = -7$$

Substitute $$\lambda = -7$$ back into the equation for Circle $$C$$:

$$x^2 + y^2 + x(-7 - 4) - y(2(-7) + 2) + 5 = 0$$

$$x^2 + y^2 - 11x + 12y + 5 = 0$$

The radius $$r$$ of this circle is calculated using $$r = \sqrt{g^2 + f^2 - c}$$:

• $$g = -\frac{11}{2}$$
• $$f = 6$$
• $$c = 5$$

$$r = \sqrt{\left(-\frac{11}{2}\right)^2 + (6)^2 - 5} = \sqrt{\frac{121}{4} + 36 - 5}$$

$$r = \sqrt{\frac{121 + 144 - 20}{4}} = \sqrt{\frac{245}{4}} = \frac{\sqrt{49 \times 5}}{2} = \frac{7\sqrt{5}}{2}$$

The diameter is $$2r$$:

$$\text{Diameter} = 2 \times \frac{7\sqrt{5}}{2} = 7\sqrt{5}$$

Final Answer:

The diameter of circle $$C$$ is $$7\sqrt{5}$$, which corresponds to Option D.