Circle And System Of Circles Question 77
Question: The equation of the circle through the points of intersection of $ x^{2}+y^{2}-1=0 $ , $ x^{2}+y^{2}-2x-4y+1=0 $ and touching the line $ x+2y=0 $ , is
[Roorkee 1989]
Options:
A) $ x^{2}+y^{2}+x+2y=0 $
B) $ x^{2}+y^{2}-x+20=0 $
C) $ x^{2}+y^{2}-x-2y=0 $
D) $ 2(x^{2}+y^{2})-x-2y=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Family of circles is $ x^{2}+y^{2}-2x-4y+1+\lambda (x^{2}+y^{2}-1)=0 $ . $ (1+\lambda )x^{2}+(1+\lambda )y^{2}-2x-4y+(1-\lambda )=0 $
$ x^{2}+y^{2}-\frac{2}{1+\lambda }x-\frac{4}{1+\lambda }y+\frac{1-\lambda }{1+\lambda }=0 $ ….. (i) Centre is $ [ \frac{1}{1+\lambda },\ \frac{2}{1+\lambda } ] $ and radius $ =\sqrt{{{( \frac{1}{1+\lambda } )}^{2}}+{{( \frac{2}{1+\lambda } )}^{2}}-\frac{1-\lambda }{1+\lambda }}=\frac{\sqrt{4+{{\lambda }^{2}}}}{1+\lambda } $ . Since it touches the line $ x+2y=0 $ ,
Hence Radius = Perpendicular from centre to the line i.e., $ | \frac{\frac{1}{1+\lambda }+2\frac{2}{1+\lambda }}{\sqrt{1^{2}+2^{2}}} |=\frac{\sqrt{4+{{\lambda }^{2}}}}{1+\lambda }\Rightarrow \sqrt{5}=\sqrt{4+{{\lambda }^{2}}}\Rightarrow \lambda \pm 1 $
$ \Rightarrow \sqrt{5}=\sqrt{4+{{\lambda }^{2}}}\Rightarrow \lambda =\pm 1 $
$ \lambda =-1 $ cannot be possible in case of circle. So $ \lambda =1 $ . Thus, from (i) $ x^{2}+y^{2}-x-2y=0 $ is the required equation of the circle.
BETA
