Let $$C: x^2 + y^2 = 4$$ and $$C': x^2 + y^2 - 4\lambda x + 9 = 0$$ be two circles. If the set of all values of $$\lambda$$ so that the circles $$C$$ and $$C'$$ intersect at two distinct points, is $$R - a, b $$, then the point $$8a + 12, 16b - 20$$ lies on the curve:

Two circles intersect at two distinct points if $$|r_1 - r_2| < d < r_1 + r_2$$, where $$d$$ is the distance between centers.

Circle Properties:

o $$C$$: Center $$C_1(0,0)$$, radius $$r_1 = 2$$.

o $$C'$$: Center $$C_2(2\lambda, 0)$$, radius $$r_2 = \sqrt{(2\lambda)^2 - 9} = \sqrt{4\lambda^2 - 9}$$.

For intersection:

$$d = \sqrt{(2\lambda - 0)^2 + 0} = |2\lambda|$$.

We need $$r_2$$ to be real: $$4\lambda^2 - 9 > 0 \implies |\lambda| > 3/2$$.

Condition: $$|r_1 - r_2| < d < r_1 + r_2 \implies (r_1 - r_2)^2 < d^2 < (r_1 + r_2)^2$$.

$$d^2 < (r_1 + r_2)^2 \implies 4\lambda^2 < (2 + \sqrt{4\lambda^2 - 9})^2$$

$$4\lambda^2 < 4 + (4\lambda^2 - 9) + 4\sqrt{4\lambda^2 - 9} \implies 5 < 4\sqrt{4\lambda^2 - 9}$$

$$25 < 16(4\lambda^2 - 9) \implies 25 < 64\lambda^2 - 144 \implies 64\lambda^2 > 169 \implies |\lambda| > 13/8$$.

Also check $$(r_1 - r_2)^2 < d^2$$, which yields the same result.

The range is $$|\lambda| > 13/8$$, which is $$\mathbb{R} - [-13/8, 13/8]$$.

So, $$a = -13/8$$ and $$b = 13/8$$.

$$x = 8a + 12 = 8(-13/8) + 12 = -13 + 12 = -1$$.

$$y = 16b - 20 = 16(13/8) - 20 = 26 - 20 = 6$$.

Point $$(x, y) = (-1, 6)$$

Plug $$(-1, 6)$$ into $$6x^2 + y^2 = 42$$: $$6(-1)^2 + (6)^2 = 6 + 36 = 42$$.

The point lies on $$6x^2 + y^2 = 42$$.