Let the 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 $\mathrm{S}$, then :

  1. $\frac{25}{9}

  2. $100<\mathrm{C}<165$

  3. $81

  4. $100<\mathrm{C}<156$

Correct Option: , 4


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

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

Centre $\equiv(-g,-f) \equiv\left(\frac{3}{2}, \frac{-10}{6}\right)$

radius $=r=\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}$


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

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

$\Rightarrow \mathrm{C}>100$ .........(1)

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

$\therefore \mathrm{S}(2,-1)<0$


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

$\mathrm{C}<156$ ..........(2)

From (1) \& (2)

$100<\mathrm{C}<156$ Ans.

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