Circle And System Of Circles Question 81
Question: The equation of the circle having as a diameter, the chord $ x-y-1=0 $ of the circle $ 2x^{2}+2y^{2}-2x-6y-25=0 $ , is
Options:
A) $ x^{2}+y^{2}-3x-y-\frac{29}{2}=0 $
B) $ 2x^{2}+2y^{2}+2x-5y-\frac{29}{2}=0 $
C) $ 2x^{2}+2y^{2}-6x-2y-21=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Required circle is $ (2x^{2}+2y^{2}-2x-6y-25)+\lambda (x-y-1)=0 $ Its centre is $ ( \frac{1-\lambda }{4},\ \frac{3-\lambda }{4} ) $ lies on line $ x-y-1=0 $ .
Hence we get the equation of circle $ 2x^{2}+2y^{2}-6x-2y-21=0 $ .
BETA
