Let and be two circles. If the set of all values of 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 : [2024]

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Solution

Q.
Let $C : x^{2} + y^{2} = 4$ and $C' : x^{2} + y^{2} - 4 λ x + 9 = 0$ be two circles. If the set of all values of $λ$ 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 : [2024]

1 $x^{2} - 4 y^{2} = 7$

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

3 $5 x^{2} - y = - 11$

4 $x^{2} + 2 y^{2} - 5 x + 6 y = 3$

Ans.
(2)

We have, $C : x^{2} + y^{2} = 4$ ... (i)

and $C' : x^{2} + y^{2} - 4 λ x + 9 = 0$

$\Rightarrow C' : {(x - 2 λ)}^{2} + y^{2} = 4 λ^{2} - 9$ ... (ii)

Radius of C, $r_{1} = 2$ and radius of C', $r_{2} = \sqrt{4 λ^{2} - 9}$

When two circles intersect at two points,

then $| r_{1} - r_{2} | < C C' < | r_{1} + r_{2} |$

$\Rightarrow$ $| 2 - \sqrt{4 λ^{2} - 9} |$ $< 2 λ < 2 + \sqrt{4 λ^{2} - 9}$ ...(iii)

By (iii), we have $4 + 4 λ^{2} - 9 - 4 \sqrt{4 λ^{2} - 9} < 4 λ^{2}$

$\Rightarrow - 5 - 4 \sqrt{4 λ^{2} - 9} < 0$

$\Rightarrow 5 + 4 \sqrt{4 λ^{2} - 9} > 0 \Rightarrow \sqrt{4 λ^{2} - 9} > - \frac{5}{4}$

On Squaring both sides, we get

$4 λ^{2} - 9 > \frac{25}{16} \Rightarrow 4 λ^{2} > \frac{169}{16} \Rightarrow λ^{2} > \frac{169}{64}$

$\Rightarrow λ > \frac{13}{8} or λ < - \frac{13}{8}$

$∴ λ \in (- \infty, - \frac{13}{8}) \cup (\frac{13}{8}, \infty)$

Also, $4 λ^{2} < {(2 + \sqrt{4 λ^{2} - 9})}^{2}$ [By (iii)]

$\Rightarrow 4 λ^{2} < 4 + 4 \sqrt{4 λ^{2} - 9} + 4 λ^{2} - 9$

$\Rightarrow 0 < - 5 + 4 \sqrt{4 λ^{2} - 9} \Rightarrow 5 < 4 \sqrt{4 λ^{2} - 9}$

On squaring, we get

$\frac{25}{16} < 4 λ^{2} - 9 \Rightarrow \frac{169}{64} < λ^{2}$

$∴ λ \in (- \infty, - \frac{13}{8}) \cup (\frac{13}{8}, \infty)$

Thus, Circles C and C' intersect at two distinct points for

$λ \in R - [\frac{- 13}{8}, \frac{13}{8}]$

$∴ a = \frac{- 13}{8}, b = \frac{13}{8}$

$∴$ (8a + 12, 16b –20) = (–1, 6) which satisfies only $6 x^{2} + y^{2} = 42$ .

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