Circle And System Of Circles Question 119
Question: If the vertices of a triangle be $ (2,\ -2) $ , $ (-1,\ -1) $ and (5, 2), then the equation of its circumcircle is
Options:
A) $ x^{2}+y^{2}+3x+3y+8=0 $
B) $ x^{2}+y^{2}-3x-3y-8=0 $
C) $ x^{2}+y^{2}-3x+3y+8=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let us find the equation of family of circles through $ (2,\ -2) $ and $ (-1,\ -1) $ . i.e. $ (x-2)(x+1)+(y+2)(y+1)+\lambda ( \frac{y+2}{-2+1}-\frac{x-2}{2+1} )=0 $
Now for point (5, 2) to lie on it, we should have $ \lambda $ given by $ 3\ .\ 6+4\ .\ 3+\lambda ( \frac{4}{-1}-1 )=0\Rightarrow \lambda =\frac{30}{5}=6 $
Hence equation is $ (x-2)(x+1)+(y+2)(y+1)+6( \frac{y+2}{-1}-\frac{x-2}{3} )=0 $ or $ x^{2}+y^{2}-3x-3y-8=0 $ .
Trick: Here the circle $ x^{2}+y^{2}-3x-3y-8=0 $ is satisfied by (2, -2), (-1, -1) and (5, 2). Therefore students must check such type of problems conversely.
BETA
