Circle And System Of Circles Question 8
Question: The values of constant term in the equation of circle passing through (1, 2) and (3, 4) and touching the line $ 3x+y-3=0 $ , is
Options:
A) 7 and 12
B) Only 7
C) Only 12
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let circle through (1, 2) and (3, 4) as ends of diameter be $ (x-1)(x-3)+(y-2)(y-4)=0 $ and line through these points is $ (y-2)=( \frac{4-2}{3-1} )(x-1) $ . Family of circles through the points and the line are $ (x-1)(x-3)+(y-2)(y-4)+\lambda (y-x-1)=0 $ .
$ \Rightarrow x^{2}+y^{2}+x(-4-\lambda )+y(-6+\lambda )+(3+8-\lambda )=0 $ This circle touches $ 3x+y-3=0 $ . Now, radius = perpendicular distance between the centre and the line.
$ \Rightarrow \sqrt{{{( \frac{4+\lambda }{2} )}^{2}}+{{( \frac{6-\lambda }{2} )}^{2}}-(3+8-\lambda )} $
$ =| \frac{3( \frac{4+\lambda }{2} )+( \frac{6-\lambda }{2} )-3}{\sqrt{5}} | $
$ \Rightarrow \lambda =1,\ -4\Rightarrow c=7 $ and 12.
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