The line $$y = x + 1$$ meets the ellipse $$\frac{x^2}{4} + \frac{y^2}{2} = 1$$ at two points $$P$$ and $$Q$$. If $$r$$ is the radius of the circle with $$PQ$$ as diameter then $$(3r)^2$$ is equal to

We need to find the intersection of the line $$y = x + 1$$ with the ellipse $$\frac{x^2}{4} + \frac{y^2}{2} = 1$$, then find $$PQ$$ as a diameter of a circle with radius $$r$$, and compute $$(3r)^2$$.

Step 1: Find the intersection points.

Substituting $$y = x + 1$$ into the ellipse equation:

$$\frac{x^2}{4} + \frac{(x+1)^2}{2} = 1$$

$$\frac{x^2}{4} + \frac{x^2 + 2x + 1}{2} = 1$$

Multiplying through by 4:

$$x^2 + 2(x^2 + 2x + 1) = 4$$

$$x^2 + 2x^2 + 4x + 2 = 4$$

$$3x^2 + 4x - 2 = 0$$

Step 2: Use Vieta's formulas.

Let the roots be $$x_1$$ and $$x_2$$. By Vieta's formulas:

$$x_1 + x_2 = -\frac{4}{3}$$ $$-(1)$$

$$x_1 x_2 = -\frac{2}{3}$$ $$-(2)$$

Step 3: Compute $$|PQ|^2$$.

Since both points lie on $$y = x + 1$$, we have $$P = (x_1, x_1 + 1)$$ and $$Q = (x_2, x_2 + 1)$$.

$$|PQ|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = (x_1 - x_2)^2 + (x_1 - x_2)^2 = 2(x_1 - x_2)^2$$

Now: $$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2 = \frac{16}{9} + \frac{8}{3} = \frac{16}{9} + \frac{24}{9} = \frac{40}{9}$$

So: $$|PQ|^2 = 2 \times \frac{40}{9} = \frac{80}{9}$$

Step 4: Find the radius.

Since $$PQ$$ is the diameter of the circle: $$|PQ| = 2r$$, so $$|PQ|^2 = 4r^2$$.

$$4r^2 = \frac{80}{9}$$

$$r^2 = \frac{80}{36} = \frac{20}{9}$$

Step 5: Compute $$(3r)^2$$.

$$(3r)^2 = 9r^2 = 9 \times \frac{20}{9} = 20$$

The answer is $$20$$, which matches Option A.