Circle And System Of Circles Question 249
Question: A circle passes through the origin and has its centre on $ y=x $ . If it cuts $ x^{2}+y^{2}-4x-6y+10=0 $ orthogonally, then the equation of the circle is
[EAMCET 1994]
Options:
A) $ x^{2}+y^{2}-x-y=0 $
B) $ x^{2}+y^{2}-6x-4y=0 $
C) $ x^{2}+y^{2}-2x-2y=0 $
D) $ x^{2}+y^{2}+2x+2y=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Let the required circle be $ x^{2}+y^{2}+2gx+2fy+c=0 $ …………. (i)
This passes through (0, 0), therefore $ c=0 $ . The centre $ (-g,\ -f) $ of (i) lies on $ y=x $ ,
Hence $ g=f $ .
Since (i) cuts the circle $ x^{2}+y^{2}-4x-6y+10=0 $ orthogonally, therefore $ 2(-2g-3f)=c+10\Rightarrow -10g=10 $
$ (\because \ g=f\ and\ c=0) $
$ \Rightarrow g=f=-1 $
Hence the required circle is $ x^{2}+y^{2}-2x-2y=0 $ .
BETA
