The set of values of k for which the circle C | vedclass

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(D) The equation of the circle is $4x^{2} + 4y^{2} - 12x + 8y + k = 0$. Dividing by $4$,we get $x^{2} + y^{2} - 3x + 2y + k/4 = 0$. Comparing with $x^

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$(9, 92/9]$

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(D) The equation of the circle is $4x^{2} + 4y^{2} - 12x + 8y + k = 0$. Dividing by $4$,we get $x^{2} + y^{2} - 3x + 2y + k/4 = 0$. Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we have $g = -3/2$,$f = 1$,$c = k/4$. The center is $(-g, -f) = (3/2, -1)$. The radius $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{9/4 + 1 - k/4} = \sqrt{(13 - k)/4} = \frac{\sqrt{13 - k}}{2}$. For the circle to exist,$r^{2} \geq 0$ $\Rightarrow 13 - k \geq 0$ $\Rightarrow k \leq 13$. $(i)$ The point $(1, -1/3)$ lies on or inside the circle,so $S(1, -1/3) \leq 0$. $1^{2} + (-1/3)^{2} - 3(1) + 2(-1/3) + k/4 \leq 0 $ $\Rightarrow 1 + 1/9 - 3 - 2/3 + k/4 \leq 0 $ $\Rightarrow k/4 \leq 3 - 1 - 1/9 + 2/3 = 2 + 5/9 = 23/9 $ $\Rightarrow k \leq 92/9$. (ii) For the circle to lie in the fourth quadrant,the distance from the center $(3/2, -1)$ to the axes must be greater than or equal to the radius. Distance to $x$-axis: $|-1| = 1$. So $r \leq 1$ $\Rightarrow \frac{\sqrt{13 - k}}{2} \leq 1$ $\Rightarrow \sqrt{13 - k} \leq 2$ $\Rightarrow 13 - k \leq 4$ $\Rightarrow k \geq 9$. Distance to $y$-axis: $|3/2| = 1.5$. So $r \leq 1.5$ $\Rightarrow \frac{\sqrt{13 - k}}{2} \leq 1.5$ $\Rightarrow \sqrt{13 - k} \leq 3$ $\Rightarrow 13 - k \leq 9$ $\Rightarrow k \geq 4$. Combining all conditions: $k \leq 92/9$ and $k \geq 9$. Thus,$k \in (9, 92/9]$.

The set of values of $k$ for which the circle $C : 4x^{2} + 4y^{2} - 12x + 8y + k = 0$ lies inside the fourth quadrant and the point $(1, -1/3)$ lies on or inside the circle $C$ is

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