JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q12

← JEE Main 2022 · 27 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 12 of 90

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 $\left(1,-\dfrac13\right)$ lies on or inside the circle C, is

an empty set

\left(6,\dfrac{65}9\right]

\left[\dfrac{80}9,10\right)

\left(9,\dfrac{92}9\right]

Answer and worked solution

Official answer: D

Circle. Centre $\left(\frac32,-1\right)$ , $r^2=\dfrac{13-k}4$ .

In the fourth quadrant. Need $r<1$ : $k>9$ .

Point on or inside. $1+\frac19-3-\frac23+\frac k4\le0$ gives $k\le\frac{92}9$ .

Set. $\left(9,\dfrac{92}9\right]$ .

(D)

Topics:Equation of a circle Inequalities and modulus

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 12