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

The Correct Option is D

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.