Let the circle S: 36 x^{2}+36 y^{2}-108 x+120 | vedclass

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(A) The given circle equation is $36 x^{2}+36 y^{2}-108 x+120 y+C=0$.Dividing by $36$,we get $x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36}=0$.The cente

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$100 < C < 156$

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(A) The given circle equation is $36 x^{2}+36 y^{2}-108 x+120 y+C=0$.Dividing by $36$,we get $x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36}=0$.The center of the circle is $(h, k) = (\frac{3}{2}, -\frac{5}{3})$.The radius $r$ is given by $r = \sqrt{h^{2}+k^{2}-\frac{C}{36}} = \sqrt{\frac{9}{4}+\frac{25}{9}-\frac{C}{36}}$.Since the circle does not intersect or touch the coordinate axes,the distance from the center to the axes must be greater than the radius.$|h| > r$ $\Rightarrow \frac{3}{2} > r$ $\Rightarrow \frac{9}{4} > \frac{9}{4}+\frac{25}{9}-\frac{C}{36}$ $\Rightarrow \frac{C}{36} > \frac{25}{9}$ $\Rightarrow C > 100$.$|k| > r$ $\Rightarrow \frac{5}{3} > r$ $\Rightarrow \frac{25}{9} > \frac{9}{4}+\frac{25}{9}-\frac{C}{36}$ $\Rightarrow \frac{C}{36} > \frac{9}{4}$ $\Rightarrow C > 81$.Combining these,we get $C > 100$.Now,the intersection of $x-2 y=4$ and $2 x-y=5$ is found by solving the system: $x=2y+4$ $\Rightarrow 2(2y+4)-y=5$ $\Rightarrow 3y=-3$ $\Rightarrow y=-1, x=2$.The point $(2, -1)$ lies inside the circle $S$,so $S(2, -1) $x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36} $5-6-\frac{10}{3}+\frac{C}{36} Thus,$100 < C < 156$.

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 coordinate axes. If the point of intersection of the lines $x-2 y=4$ and $2 x-y=5$ lies inside the circle $S$,then :

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