The set of values of k, for which the circle C:4x2+4y2−12x+8y+k=0 lies inside the fourth quadrant and the point \[\left( {1, - {1 \over 3}} \right)\] lies on or inside the circle C, is :
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Official solution
C:4x2+4y2−12x+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 4th quadrant centre \[ \equiv \left( {{3 \over 2}, - 1} \right)\]
\[\therefore\] \[r {9 \over 4}\]
\[ \Rightarrow k > 9\]
\[\therefore\] \[k \in \left( {9,{{92} \over 9}} \right)\]