Question

Let the circle $S:36x^2+36y^2-108x+120y+C=0$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $x-2y=4$ and $2x-y=5$ lies inside the circle $S$, then:

• A. $\frac{25}{9}<C<\frac{13}{3}$
• B. $81<C<156$
• C. $100<C<156$
• D. $100<C<165$

Answer 1

The correct option is C $100<C<156$

Intersection point of $x-2y=4$ and $2x-y=5$ is $(2,-1).$
$(2,-1)$ lies inside the given circle

$\therefore 36\left(4\right)+36\left(1\right)-108\left(2\right)+120(-1)+C<0\Rightarrow c<156\cdots \left(i\right)$
and ${\left(\frac{108}{72}\right)}^2+{\left(\frac{-120}{72}\right)}^2-\frac{C}{36}<{\left(\frac{3}{2}\right)}^2$ (Neither touches any axis)
$\Rightarrow \frac{9}{4}+\frac{25}{9}-\frac{C}{36}<\frac{9}{4}$
$\therefore 100<C\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$ $100<C<156$