QJEEM22Maf6c0: JEE Main 2022 Mathematics - Class 11 JEEM22Maf6c0 Solution: The set of values of k, for which the circle lies inside the…

$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$

$\because$ $\left( {1, - {1 \over 3}} \right)$ lies on or inside the C

then $4 + {4 \over 9} - 12 - {8 \over 3} + k \le 0$

$\Rightarrow k \le {{92} \over 9}$

Now, circle lies in 4^th quadrant centre $\equiv \left( {{3 \over 2}, - 1} \right)$

$\therefore$ r < 1 \Rightarrow \sqrt {{9 \over 4} + 1 - {k \over 4}} < 1

\Rightarrow {{13} \over 4} - {k \over 4} < 1

\Rightarrow {k \over 4} > {9 \over 4}

\Rightarrow k > 9

$\therefore$ $k \in \left( {9,{{92} \over 9}} \right)$