JEE Main 2021 — Circle Question with Solution

The given circle is $S:36x^2+36y^2-108x+120y+C=0$

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

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

Thus, centre $e\equiv \left(-g,-f\right)\equiv \left(\frac{3}{2}, -\frac{10}{6}\right)$ and radius $=r=\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}$

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.

$\Rightarrow r<\frac{3}{2}$

$\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}<\frac{3}{2}$

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

$\Rightarrow C>100 \dots \left(1\right)$

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

The point $\left(x_1, y_1\right)$ lies inside a circle $x^2+y^2+2gx+2fy+c=0,$ if $x_1^2+y_1^2+2g{x}_1+2f{y}_1+c<0$

$\Rightarrow {\left(2\right)}^2+{\left(-1\right)}^2-3\left(2\right)+\frac{10}{3}\left(-1\right)+\frac{C}{36}<0$

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

$\Rightarrow C<156 \dots \left(2\right)$

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