JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Circle

Let the circle

S : 36 x^{2} + 36 y^{2} - 108 x + 120 y + C = 0

be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines,

x - 2 y = 4

and

2 x - y = 5

lies inside the circle S, then :

Answer: (C)

Solution

Dividing by

36

gives

x^{2} + y^{2} - 3 x + \frac{10}{3} y + \frac{C}{36} = 0

, a circle with centre

(\frac{3}{2}, - \frac{5}{3})

and radius

r = \frac{9}{4} + \frac{25}{9} - \frac{C}{36}

. Missing both axes means

r

is smaller than each distance from the centre to an axis, namely

\frac{3}{2}

and

\frac{5}{3}

; the tighter of these is

r < \frac{3}{2}

, which gives

\frac{9}{4} + \frac{25}{9} - \frac{C}{36} < \frac{9}{4} \Rightarrow C > 100

The lines

x - 2 y = 4

and

2 x - y = 5

meet at

(2, - 1)

, and a point lies inside the circle when the left side of the circle's equation is negative there:

36 (4) + 36 (1) - 108 (2) + 120 (- 1) + C < 0 \Rightarrow C < 156

Both requirements together give

100 < C < 156

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option