Test Index

Circles Part 2

Section: Mathematics

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Question : 38 of 87

Marks: +1, -0

Let

C: x^2+y^2=4

C^{'}: x^2+y^2-4\lambda x+9=0

be two circles. If the set of all values of

\lambda

so that the circles

C

C^{'}

intersect at two distinct points, is

R - [a, b]

, then the point

(8a+12,16b-20)

lies on the curve :

[1-Feb-2024 Shift 1]

$x^2+2y^2-5x+6y=3$
$5x^2-y=-11$
$x^2-4y^2=7$
$6x^2+y^2=42$

Solution: 👈: Video Solution

\;x^2+y^2=4

\; C(0,0)

\; r_1=2

\; C^{'}(2\lambda, 0) \;\; r_2=\sqrt{4\lambda^2-9}

\;|r_1-r_2| < CC^{'}<|r_1+r_2|

\;2-\sqrt{4\lambda^2-9}|<|2\lambda|

<2+\sqrt{4\lambda^2-9}

\;4+4\lambda^2-9-4\sqrt{4\lambda^2-9}<4\lambda^2

True

\lambda \in

R... (1)

\;4\lambda^2<4+4\lambda^2-9+4\sqrt{4\lambda^2-9}

\;5<4\sqrt{4\lambda^2-9}\;\text{ and }\;\;\lambda^2\ge\frac{9}{4}

\;\;\frac{25}{16}<4\lambda^2-9\;\;\;\;\;

\lambda \in\left(-\infty,-\frac{3}{2}]\cup[\frac{3}{2},\infty\right)

\;\;\frac{169}{64} < \lambda^2

\lambda \in\left(-\infty,-\frac{13}{8}\right)\cup\left(\frac{13}{8},\infty\right)

........(2)

from (1) and (2)

\lambda \in

\lambda \in\left(-\infty,-\frac{13}{8}\right)\cup\left(\frac{13}{8},\infty\right)

\Rightarrow R-\left[-\frac{13}{8},\frac{13}{8}\right]

as per question

a=-\frac{13}{8}

and

b=\frac{13}{8}

required point is

(-1,6)

with satisfies option (4)

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