Circle And System Of Circles Question 244
The equation of the circle having the lines $ x^{2}+2xy+3x+6y=0 $ as its tangents and having size just sufficient to contain the circle $ x(x-4)+y(y-3)=0 $ is
[Roorkee 1990]
Options:
A) $ x^{2}+y^{2}+3x-6y-40=0 $
B) $ x^{2}+y^{2}+6x-3y-45=0 $
C) $ x^{2}+y^{2}+8x+4y-20=0 $
D) $ x^{2}+y^{2}+4x+8y+20=0 $
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Answer:
Correct Answer: B
Solution:
Given circle is $ ( 2,\ \frac{3}{2} ),\ r_1=\frac{5}{2} $ (say)
Required normals of circles are $ x+3=0,\ x+2y=0 $ which intersect at the centre $ ( -3,\ \frac{3}{2} ),\ r_2= $ radius (say).
2nd circle just contains the 1st i.e., $ C_2C_1=r_1-r_2\Rightarrow r_2=\frac{15}{2} $ .
BETA
