Test Index

Circles Part 1

Section: Mathematics

© examsnet.com

Question : 14 of 100

Marks: +1, -0

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

2x-y=5

lies inside the circle

S

[22 Jul 2021 Shift 2]

$\frac{25}{9} < C < \frac{13}{3}$
$100 < C < 165$
$81 < C < 156$
$100 < C < 156$

Solution:

S: 36x^{2}+36y^{2}-108x+120y

+C=0

\Rightarrow x^{2}+y^{2}-3x+\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}}

Now,

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

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

\Rightarrow C>100

........(1)

Now point of intersection of

x-2y=4

and

2x-y=5

is

(2,-1)

, which lies inside the circle

S

.

\therefore S(2,-1) < 0

\Rightarrow (2)^{2}+(-1)^{2}-3(2)+\frac{10}{3}(-1)

+\frac{C}{36} < 0

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

C < 156

..........(2) From

(1) \& (2)

100 < C < 156

Ans.

© examsnet.com

Go to Question:

Prev Question

Next Question