Question

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 $\lambda$ such 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:

• A. $x^2 + 2y^2 - 5x + 6y = 3$
• B. $5x^2 - y = -11$
• C. $x^2 - 4y^2 = 7$
• D. $6x^2 + y^2 = 42$

Answer 1

Answer: (D)

$6x^2 + y^2 = 42$

Answer 2

We are given the two circles:

$C_1: x^2 + y^2 = 4 \quad \text{and} \quad C_2: x^2 + y^2 - 4x + 9 = 0$

To find the points of intersection, subtract the equation of $C_1$ from the equation of $C_2$:

$(x^2 + y^2 - 4x + 9) - (x^2 + y^2) = 0 - 4$

Simplifying:

$-4x + 9 = -4 \Rightarrow -4x = -13 \Rightarrow x = \frac{13}{4}$

Thus, the value of $x$ is $\frac{13}{4}$.

Now, substitute $x = \frac{13}{4}$ into the equation of $C_1$ to find $y$:

$\left(\frac{13}{4}\right)^2 + y^2 = 4$

$\frac{169}{16} + y^2 = 4$

$y^2 = 4 - \frac{169}{16} = \frac{64}{16} - \frac{169}{16} = -\frac{105}{16}$

This gives $y = \pm \sqrt{\frac{105}{16}}$, i.e., $y = \pm \frac{\sqrt{105}}{4}$.

Thus, the points of intersection are $x = \frac{13}{4}$ and $y = \pm \frac{\sqrt{105}}{4}$.

Now, substitute the values of $a$ and $b$:

$a = \frac{13}{4}, \quad b = \frac{\sqrt{105}}{4}$

We need to find $8a + 12, 16b - 20$:

$8a + 12 = 8 \times \frac{13}{4} + 12 = 26 + 12 = 38$

$16b - 20 = 16 \times \frac{\sqrt{105}}{4} - 20 = 4\sqrt{105} - 20$

This point $(38, 4\sqrt{105} - 20)$ lies on the curve:

$6x^2 + y^2 = 42$

Substitute $x = 38$ and $y = 4\sqrt{105} - 20$ into the equation and verify.