Circle And System Of Circles Question 60
Question: The equation of a circle passing through the point (4, 5) and having the centre at (2, 2) is
[UPSEAT 2000]
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:
Given that centre of circle is (2, 2) Since circle is passing from point (4, 5) So, radius of circle $ =\sqrt{{{(4-2)}^{2}}+{{(5-2)}^{2}}} $
$ =\sqrt{(4+9)}=\sqrt{13} $ Therefore, equation of circle is $ {{(x-2)}^{2}}+{{(y-2)}^{2}}={{(\sqrt{13})}^{2}} $ or $ x^{2}-4x+4+y^{2}-4y+4=13 $ or $ x^{2}+y^{2}-4x-4y-5=0 $ .
BETA
