Circle And System Of Circles Question 130
Question: The equation of the circle having centre $ (1,\ -2) $ and passing through the point of intersection of lines $ 3x+y=14 $ , $ 2x+5y=18 $ is
[MP PET 1990]
Options:
A) $ x^{2}+y^{2}-2x+4y-20=0 $
B) $ x^{2}+y^{2}-2x-4y-20=0 $
C) $ x^{2}+y^{2}+2x-4y-20=0 $
D) $ x^{2}+y^{2}+2x+4y-20=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
The point of intersection of $ 3x+y-14=0 $ and $ 2x+5y-18=0 $ are $ x=\frac{-18+70}{15-2},\ y=\frac{-28+54}{13}\Rightarrow x=4,\ y=2 $ i.e., point is (4, 2).
Therefore radius is $ \sqrt{(9)+(16)}=5 $ and equation is $ x^{2}+y^{2}-2x+4y-20=0 $ .
Trick : The only circle is $ x^{2}+y^{2}-2x+4y-20=0 $ , whose centre is (1, -2).
BETA
