Circle And System Of Circles Question 136
Question: The equation of circle passing through (4, 5) and having the centre at (2, 2), is
[MNR 1986; MP PET 1984]
Options:
A) $ x^{2}+y^{2}+4x+4y-5=0 $
B) $ x^{2}+y^{2}-4x-4y-5=0 $
C) $ x^{2}+y^{2}-4x=13 $
D) $ x^{2}+y^{2}-4x-4y+5=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Centre (2, 2) and $ R=\sqrt{{{(4-2)}^{2}}+{{(5-2)}^{2}}}=\sqrt{13} $
Hence required equation is $ x^{2}+y^{2}-4x-4y-5=0 $ .
BETA
