Circle And System Of Circles Question 25
Question: Given the circles $ x^{2}+y^{2}-4x-5=0 $ and $ x^{2}+y^{2}+6x-2y+6=0 $ . Let P be a point $ (\alpha ,\beta ) $ such that the tangents from P to both the circles are equal, then
Options:
A) $ 2\alpha +10\beta +11=0 $
B) $ 2\alpha -10\beta +11=0 $
C) $ 10\alpha -2\beta +11=0 $
D) $ 10\alpha +2\beta +11=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Accordingly, $ {{\alpha }^{2}}+{{\beta }^{2}}-4\alpha -5={{\alpha }^{2}}+{{\beta }^{2}}+6\alpha -2\beta +6 $
$ \Rightarrow 10\alpha -2\beta +11=0 $ .
BETA
