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The radius of the circle having the lines \mathrm{x^2+2 x y+3 x+6 y=0} as its normal and having size just sufficient to contain the circle \mathrm{x(x-4)+y(y-3)=0} _________.

Option: 1

(7.5)


Option: 2

(7.3)


Option: 3

(3.3)


Option: 4

(4.3)


Answers (1)

best_answer

The equation x^2+2 x y+3 x+6 y=0 can be written on factorizing the L.H.S, as

(x+3)(x+2 y)=0 .
Thus the equations of the two normals to the required circle are x+3=0 \, \, and \, \, x+2 y=0.

Solving these, we get the co-ordinates of the centre as (-3,3 / 2). Again as the required circle just contains the circle

x(x-4)+y(y-3)=0, that is

x^2+y^2-4 x-3 y=0                              ...(1)

This means that the required circle will touch the circle (1) internally. Let r be the radius of required circle. Also the radius of

the circle ( 1) is \sqrt{2^2+(3 / 2)^2-0}=5 / 2 and its centre is (2,3 / 2).

Now the required circle will touch the circle (1) internally, if the distance between their centres is equal to the difference of their

radii that is,

\sqrt{ }\left[(-3-2)^2+(3 / 2-3 / 2)^2\right]=r-5 / 2

\Rightarrow 5=r-5 / 2$, giving $r=15 / 2.

Posted by

Divya Prakash Singh

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