Circle And System Of Circles Question 340
Question: If O is the origin and OP, OQ are tangents to the circle $ x^{2}+y^{2}+2gx+2fy+c=0 $ , the circumcentre of the triangle $ OPQ $ is
Options:
A) $ (-g,,-f) $
B) $ (g,f) $
C) $ (-f,-g) $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Equation of pair of tangents from (0, 0) to circle are $ SS_1=T^{2} $ . Equation of circle through origin and chord of contact is $ x^{2}+y^{2}+2gx+2fy+c+\lambda (gx+fy+c)=0 $
$ \Rightarrow \lambda =-1 $ (by $ x=0,\ y=0 $ ) Therefore, equation is $ x^{2}+y^{2}+gx+fy=0 $ .
Hence circumcentre is $ ( -\frac{g}{2},\ -\frac{f}{2} ) $ Aliter: Required circumcentre is the mid-point of (0, 0) and $ (-g,\ -f) $ i.e., $ ( -\frac{g}{2},\ -\frac{f}{2} ) $ .
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