The set of values of k, for which the circle $$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$ lies inside the fourth quadrant and the point $$\left( {1, - {1 \over 3}} \right)$$ lies on or inside the circle C, is :
JEE · Math · previous-year question
- A.an empty set
- B.$$\left( {6,{{65} \over 9}} \right]$$
- C.$$\left[ {{{80} \over 9},10} \right)$$
- D.$$\left( {9,{{92} \over 9}} \right]$$correct
Answer
D. $$\left( {9,{{92} \over 9}} \right]$$
Explanation
$$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$ $$\because$$ $$\left( {1, - {1 \over 3}} \right)$$ 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 4th quadrant centre $$ \equiv \left( {{3 \over 2}, - 1} \right)$$ $$\therefore$$ $$r $$ \Rightarrow {{13} \over 4} - {k \over 4} $$ \Rightarrow {k \over 4} > {9 \over 4}$$ $$ \Rightarrow k > 9$$ $$\therefore$$ $$k \in \left( {9,{{92} \over 9}} \right)$$
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