Question

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 $\bigg(1, -\frac{1}{3}\bigg)$ lies on or inside the circle $C$, is

• A. An empty set
• B. $\bigg(6/\frac{65}{9}\bigg)$
• C. $\bigg(\frac{80}{9},10\bigg)$
• D. $\bigg(9,\frac{92}{9}\bigg)$

Answer 1

Answer: (D)

$\bigg(9,\frac{92}{9}\bigg)$

Answer 2

$C:4x^2+4y^2−12x+8y+k=0$

$∵(1,−13)$

lies on or inside the $C$ then

$4+\frac{4}{9}−12−\frac{8}{3}+k≤0$

$⇒k≤\frac{92}{9}$

Now, circle lies in $4^{th}$ quadrant centre $≡\bigg(\frac{3}{2},−1\bigg)$

$∴r<1⇒\sqrt{\frac{9}{4}+1−\frac{k}{4}}<1$

$⇒\frac{13}{4}−\frac{k}{4}<1$

$⇒\frac{k}{4}>\frac{9}{4}$

$⇒k > 9$

$∴k∈\bigg(9,\frac{92}{9}\bigg)$