QJEEM21Ma2a83: JEE Main 2021 Mathematics - Class 11 JEEM21Ma2a83 Solution: Let the circle S : 36x 2 + 36y 2 108x + 120y + C = 0 be such…

S : 36x² + 36y² $-$ 108x + 120y + C = 0

$\Rightarrow$ x² + y² $-$ 3x + ${{10} \over 3}$ y + ${C \over {36}}$ = 0

Centre $\equiv ( - g, - f) \equiv \left( {{3 \over 2},{{ - 10} \over 6}} \right)$

radius = $r = \sqrt {{9 \over 4} + {{100} \over {36}} - {C \over {36}}}$


Now,

\Rightarrow r < {3 \over 2}

\Rightarrow {9 \over 4} + {{100} \over {36}} - {C \over {36}} < {9 \over 4}

$\Rightarrow$ C > 100 ...... (1)

Now, point of intersection of x $-$ 2y = 4 and 2x $-$ y = 5 is (2, $-$ 1), which lies inside the circle S.

$\therefore$ S(2, $-$ 1) < 0

$\Rightarrow$ (2)² + ( $-$ 1)² $-$ 3(2) + ${{10} \over 3}$ ( $-$ 1) + ${C \over {36}}$ < 0

$\Rightarrow$ 4 + 1 $-$ 6 $-$ ${{10} \over 3}$ + ${C \over {36}}$ < 0

C < 156 ..... (2)

From (1) & (2)

100 < C < 156 Ans.