Solution
### Related Formula
For two circles with centers C_1, C_2$C_1, C_2$ and radii r_1, r_2$r_1, r_2$ to intersect at two distinct real points, the distance between their centers must satisfy:
|r_1 - r_2| < C_1C_2 < r_1 + r_2$$|r_1 - r_2| < C_1C_2 < r_1 + r_2$$
### Core Logic
Extracting properties from the circle equations:
- Circle C$C$: x^2 + y^2 = 4 implies$x^2 + y^2 = 4 \implies$ Center C_1 = (0,0)$C_1 = (0,0)$, Radius r_1 = 2$r_1 = 2$
- Circle C'$C'$: x^2 + y^2 - 4lambda x + 9 = 0 implies$x^2 + y^2 - 4\lambda x + 9 = 0 \implies$ Center C_2 = (2lambda, 0)$C_2 = (2\lambda, 0)$, Radius r_2 = sqrt(-2lambda)^2 - 9 = sqrt4lambda^2 - 9$r_2 = \sqrt{(-2\lambda)^2 - 9} = \sqrt{4\lambda^2 - 9}$
For real intersection conditions, the radius must be defined:
4lambda^2 - 9 > 0 implies lambda^2 > frac94 quad implies (1)$$4\lambda^2 - 9 > 0 \implies \lambda^2 > \frac{9}{4} \quad \implies (1)$$
Distance between centers: C_1C_2 = sqrt(2lambda - 0)^2 + 0^2 = |2lambda|$C_1C_2 = \sqrt{(2\lambda - 0)^2 + 0^2} = |2\lambda|$.
### Step 1: Formulate the Triangle Inequality Solutions
Applying the intersection condition:
|2 - sqrt4lambda^2 - 9| < |2lambda| < 2 + sqrt4lambda^2 - 9$$|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$$
The right side inequality |2lambda| < 2 + sqrt4lambda^2 - 9$|2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$ is always valid for real radii. Solving the left side inequality by squaring:
left(2 - sqrt4lambda^2 - 9right)^2 < (2lambda)^2$$\left(2 - \sqrt{4\lambda^2 - 9}\right)^2 < (2\lambda)^2$$
4 + (4lambda^2 - 9) - 4sqrt4lambda^2 - 9 < 4lambda^2$$4 + (4\lambda^2 - 9) - 4\sqrt{4\lambda^2 - 9} < 4\lambda^2$$
-5 - 4sqrt4lambda^2 - 9 < 0 implies 4sqrt4lambda^2 - 9 > -5$$-5 - 4\sqrt{4\lambda^2 - 9} < 0 \implies 4\sqrt{4\lambda^2 - 9} > -5$$
Since a square root is always non-negative, square both sides to get the strict bound:
16(4lambda^2 - 9) > 25 implies 64lambda^2 - 144 > 25$$16(4\lambda^2 - 9) > 25 \implies 64\lambda^2 - 144 > 25$$
64lambda^2 > 169 implies lambda^2 > frac16964 quad implies (2)$$64\lambda^2 > 169 \implies \lambda^2 > \frac{169}{64} \quad \implies (2)$$
Combining bounds (1) and (2), condition (2) is more restrictive, meaning:
lambda in left(-infty, -frac138right) cup left(frac138, inftyright)$$\lambda \in \left(-\infty, -\frac{13}{8}\right) \cup \left(\frac{13}{8}, \infty\right)$$
This can be written as mathbbR - left[-frac138, frac138right]$\mathbb{R} - \left[-\frac{13}{8}, \frac{13}{8}\right]$.
### Step 2: Coordinate Analysis of the Target Point
Comparing our interval with mathbbR - [a, b]$\mathbb{R} - [a, b]$, we identify:
a = -frac138, quad b = frac138$$a = -\frac{13}{8}, \quad b = \frac{13}{8}$$
Now, substitute these bounds to locate the coordinates of our target point (8a+12, \, 16b-20)$(8a+12, \, 16b-20)$:
- x$x$-coordinate: 8left(-frac138right) + 12 = -13 + 12 = -1$8\left(-\frac{13}{8}\right) + 12 = -13 + 12 = -1$
- y$y$-coordinate: 16left(frac138right) - 20 = 26 - 20 = 6$16\left(\frac{13}{8}\right) - 20 = 26 - 20 = 6$
Hence, the point is (-1, 6)$(-1, 6)$.
### Step 3: Test Point against the Options
Substitute (-1, 6)$(-1, 6)$ into the given curves to find a match:
Testing Option (4): 6x^2 + y^2 = 42$6x^2 + y^2 = 42$
6(-1)^2 + (6)^2 = 6(1) + 36 = 42$$6(-1)^2 + (6)^2 = 6(1) + 36 = 42$$
Since LHS = RHS, the point satisfies the curve in Option (4).
### Pattern Recognition
Sees: Dynamic variable interval limits forming loci requirements.
Shortcut: When checking inequalities like |2 - sqrtz| < 2lambda$|2 - \sqrt{z}| < 2\lambda$, algebraic squaring helps simplify the terms quickly by eliminating the parameter 4lambda^2$4\lambda^2$ from both sides.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Conic Sections (Circles)
Class 11 Mathematics: Linear Inequalities
