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 :
Solution
<p>$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$</p>
<p>$\because$ $\left( {1, - {1 \over 3}} \right)$ lies on or inside the C</p>
<p>then $4 + {4 \over 9} - 12 - {8 \over 3} + k \le 0$</p>
<p>$\Rightarrow k \le {{92} \over 9}$</p>
<p>Now, circle lies in 4<sup>th</sup> quadrant centre $\equiv \left( {{3 \over 2}, - 1} \right)$</p>
<p>$\therefore$ $r < 1 \Rightarrow \sqrt {{9 \over 4} + 1 - {k \over 4}} < 1$</p>
<p>$\Rightarrow {{13} \over 4} - {k \over 4} < 1$</p>
<p>$\Rightarrow {k \over 4} > {9 \over 4}$</p>
<p>$\Rightarrow k > 9$</p>
<p>$\therefore$ $k \in \left( {9,{{92} \over 9}} \right)$</p>
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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