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Step-by-Step Solution
Step 1: Identify the Circles and Their Centers
The first circle is given by
$x^2 + y^2 = 4.$
Its center is $(0, 0)$, and radius $r_1 = 2.$
The second circle is given by
$x^2 + y^2 - 4\lambda x + 9 = 0.$
Rewrite
$x^2 - 4\lambda x$
as
$(x - 2\lambda)^2 - (2\lambda)^2,$
so the circle becomes
$(x - 2\lambda)^2 + y^2 = 4\lambda^2 - 9.$
Hence, it has center $(2\lambda,\,0)$ and radius
$\sqrt{4\lambda^2 - 9},$
which is real only when
$4\lambda^2 - 9 \ge 0 \implies |\lambda| \ge \tfrac{3}{2}.$
Step 2: Condition for Two Circles to Intersect at Two Distinct Points
Two circles with centers $O_1$ and $O_2$ and radii $r_1$ and $r_2$ intersect at two distinct points if
$|r_1 - r_2| < d < r_1 + r_2,$
where $d$ is the distance between the centers.
Here:
Center of $C$: $(0,0)$, radius $r_1 = 2.$
Center of $C': (2\lambda, 0)$, radius $r_2 = \sqrt{4\lambda^2 - 9}.$
Distance between centers: $d = |2\lambda|.$
The condition becomes
$|\,2 - \sqrt{4\lambda^2 - 9}\,| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9},$
with the constraint
$|\lambda|\ge \tfrac{3}{2}$
so that $r_2$ is real.
Step 3: Solve the Inequalities
Focus on
$|2\lambda| < 2 + \sqrt{4\lambda^2 - 9}.$
Write $|2\lambda| = 2|\lambda|.$
If $2|\lambda| - 2 \ge 0$ (i.e., $|\lambda| \ge 1$), we get
$2|\lambda| - 2 < \sqrt{4\lambda^2 - 9}.$
Square both sides (valid under the noted condition):
$(2|\lambda| - 2)^2 < 4\lambda^2 - 9.$
This simplifies to:
$4(\lambda^2 - 2|\lambda| + 1) < 4\lambda^2 - 9 \implies -8|\lambda| + 4 < -9 \implies 8|\lambda| > 13 \implies |\lambda| > \tfrac{13}{8}.$
Conclusively, combining all conditions, the values of $\lambda$ for two distinct intersections must satisfy
$|\lambda| > \tfrac{13}{8}.$
Step 4: Express the Interval for $\lambda$
Therefore, $\lambda$ lies outside the closed interval
$\left[-\tfrac{13}{8}, \tfrac{13}{8}\right].$
In set-builder form,
$\lambda \in
\mathbb{R} \setminus \bigl[-\tfrac{13}{8},\,\tfrac{13}{8}\bigr].$
Hence in the problem’s notation, $a = -\tfrac{13}{8}$ and $b = \tfrac{13}{8}.$
Step 5: Find the Point $(8a + 12,\,16b - 20)$
Substitute $a = -\tfrac{13}{8}$ and $b = \tfrac{13}{8}.$
$8a = 8 \times \left(-\tfrac{13}{8}\right) = -13 \quad \Longrightarrow \quad 8a + 12 = -13 + 12 = -1.$
$16b = 16 \times \tfrac{13}{8} = 2 \times 13 = 26 \quad \Longrightarrow \quad 16b - 20 = 26 - 20 = 6.$
So the point is $(-1,\,6).$
Step 6: Verify Which Curve This Point Lies On
The claimed correct curve is
$6x^2 + y^2 = 42.$
Substitute $x = -1$ and $y = 6$:
$6 \times (-1)^2 + (6)^2 = 6 \times 1 + 36 = 6 + 36 = 42.$
This exactly satisfies
$6 x^2 + y^2 = 42,$
confirming the correct answer is indeed
$6x^2 + y^2 = 42.$
Final Answer
The point $\bigl(8a + 12,\,16b - 20\bigr) = (-1,\,6)$ lies on the curve
$6x^2 + y^2 = 42.$
