JEE Main 2021 — Mathematics Coordinate Geometry

The given circle is $S:36{x}^{2}+36{y}^{2}-108x+120y+C=0$S:36x2+36y2−108x+120y+C=0

$$\Rightarrow {x}^{2}+{y}^{2}-3x+\frac{10}{3}y+\frac{C}{36}=0$$⇒x2+y2−3x+310​y+36C​=0

We know that the centre and radius of a circle ${x}^{2}+{y}^{2}+2gx+2fy+c=0$x2+y2+2gx+2fy+c=0 is $(-g,-f)$(−g,−f) and $\sqrt{{g}^{2}+{f}^{2}-c}$g2+f2−c​ respectively.

Thus, centre $e\equiv (-g,-f)\equiv (\frac{3}{2},-\frac{10}{6})$e≡(−g,−f)≡(23​,−610​) and radius $=r=\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}$=r=49​+36100​−36C​​

Now, this circle neither intersects nor touches the co-ordinate axes, hence, the radius of the circle is less than the absolute value of the co-ordinate with smaller value.

$$\Rightarrow r<\frac{3}{2}$$⇒r<23​

$$\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}<\frac{3}{2}$$49​+36100​−36C​​<23​

$$\Rightarrow \frac{9}{4}+\frac{100}{36}-\frac{C}{36}<\frac{9}{4}$$⇒49​+36100​−36C​<49​

$$\Rightarrow C>100\ldots (1)$$⇒C>100…(1)

Now, point of intersection of $x-2y=4$x−2y=4 and $2x-y=5$2x−y=5 is $(2,-1),$(2,−1), which lies inside the circle $S.$S.

The point $({x}_{1},{y}_{1})$(x1​,y1​) lies inside a circle ${x}^{2}+{y}^{2}+2gx+2fy+c=0,$x2+y2+2gx+2fy+c=0, if ${x}_{1}^{2}+{y}_{1}^{2}+2g{x}_{1}+2f{y}_{1}+c<0$x12​+y12​+2gx1​+2fy1​+c<0

$$\Rightarrow {(2)}^{2}+{(-1)}^{2}-3(2)+\frac{10}{3}(-1)+\frac{C}{36}<0$$⇒(2)2+(−1)2−3(2)+310​(−1)+36C​<0

$$\Rightarrow 4+1-6-\frac{10}{3}+\frac{C}{36}<0$$⇒4+1−6−310​+36C​<0

$$\Rightarrow C<156\ldots (2)$$⇒C<156…(2)

From $(1)$(1) and $(2),$(2), we have $100<C<156$100<C<156