Statement 1: The only circle having radius $$\sqrt{10}$$ and a diameter along line $$2x + y = 5$$ is $$x^2 + y^2 - 6x + 2y = 0$$.
Statement 2: $$2x + y = 5$$ is a normal to the circle $$x^2 + y^2 - 6x + 2y = 0$$.

First, we need to evaluate Statement 1 and Statement 2.

Statement 1 claims that the only circle with radius $$\sqrt{10}$$ and a diameter along the line $$2x + y = 5$$ is given by the equation $$x^2 + y^2 - 6x + 2y = 0$$.

Statement 2 claims that the line $$2x + y = 5$$ is a normal to the circle $$x^2 + y^2 - 6x + 2y = 0$$.

We start by verifying Statement 2. A normal to a circle passes through its center. So, we need to find the center of the circle $$x^2 + y^2 - 6x + 2y = 0$$. To do this, we rewrite the equation in standard form by completing the square.

For the x-terms: $$x^2 - 6x$$ can be written as $$(x-3)^2 - 9$$.

For the y-terms: $$y^2 + 2y$$ can be written as $$(y+1)^2 - 1$$.

Substituting these into the equation:

$$(x-3)^2 - 9 + (y+1)^2 - 1 = 0$$

Simplifying:

$$(x-3)^2 + (y+1)^2 - 10 = 0$$

So,

$$(x-3)^2 + (y+1)^2 = 10$$

This shows that the circle has center $$(3, -1)$$ and radius $$\sqrt{10}$$.

Now, we check if the line $$2x + y = 5$$ passes through the center $$(3, -1)$$. Substituting $$x = 3$$ and $$y = -1$$:

$$2(3) + (-1) = 6 - 1 = 5$$

This equals 5, so the point lies on the line. Therefore, the line passes through the center, meaning it is a normal to the circle. Hence, Statement 2 is true.

Next, we evaluate Statement 1. It states that the circle $$x^2 + y^2 - 6x + 2y = 0$$ is the only circle with radius $$\sqrt{10}$$ and a diameter along the line $$2x + y = 5$$.

For a circle to have a diameter along a line, the center must lie on that line because the center is the midpoint of the diameter. Additionally, the radius is given as $$\sqrt{10}$$.

So, any circle with center on the line $$2x + y = 5$$ and radius $$\sqrt{10}$$ will have a diameter along this line. Let the center be $$(h, k)$$. Then, it must satisfy the line equation:

$$2h + k = 5$$

The general equation of such a circle is:

$$(x - h)^2 + (y - k)^2 = (\sqrt{10})^2 = 10$$

We know that the circle $$x^2 + y^2 - 6x + 2y = 0$$ has center $$(3, -1)$$ and radius $$\sqrt{10}$$, and it satisfies $$2(3) + (-1) = 6 - 1 = 5$$, so it is one such circle.

However, are there others? We can choose different points on the line $$2x + y = 5$$. For example:

• Let $$h = 0$$. Then $$2(0) + k = 5$$ gives $$k = 5$$. The circle equation is $$(x - 0)^2 + (y - 5)^2 = 10$$, which simplifies to $$x^2 + y^2 - 10y + 25 = 10$$, or $$x^2 + y^2 - 10y = -15$$.
• Let $$h = 1$$. Then $$2(1) + k = 5$$ gives $$k = 3$$. The circle equation is $$(x - 1)^2 + (y - 3)^2 = 10$$, which simplifies to $$x^2 - 2x + 1 + y^2 - 6y + 9 = 10$$, or $$x^2 + y^2 - 2x - 6y = 0$$.

These are two different circles with radius $$\sqrt{10}$$ and center on the line $$2x + y = 5$$, so they have a diameter along this line. Therefore, there are infinitely many such circles, not just one. Hence, Statement 1 is false because it claims that the given circle is the only one.

In summary, Statement 1 is false and Statement 2 is true. Looking at the options:

A. Statement 1 is false; Statement 2 is true.

B. Statement 1 is true; Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

C. Statement 1 is true; Statement 2 is false.

D. Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Hence, the correct answer is Option A.