Circle-And-System-Of-Circles Question 388
Question: The locus of a point which moves so that the ratio of the length of the tangents to the circles $ x^{2}+y^{2}+4x+3=0 $ and $ x^{2}+y^{2}-6x+5=0 $ is 2:3 is
[Kerala (Engg.) 2005]
Options:
A) 5, -35
B) -5, 35
C) 7, -32
D) -7, 32
E) 3, -13
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of circle is, $ x^{2}+y^{2}-2x+6y-6=0 $ $ {{(x-1)}^{2}}+{{(y+3)}^{2}}={{(4)}^{2}} $ Radius of circle = 4 And centre of circle $ =(1,-3) $ Equation of tangent $ 3x-4y+k=0 $ \ $ \frac{3\times 1-4\times (-3)+k=0}{\sqrt{{{(3)}^{2}}+{{(-4)}^{2}}}}=\pm 4 $ . Hence, $ k=5,-35 $ .
BETA
