Circle And System Of Circles Question 74
Question: If the equation of the tangent to the circle $ x^{2}+y^{2}-2x+6y-6=0 $ parallel to $ 3x-4y+7=0 $ is $ 3x-4y+k=0 $ , then the values of k are
[Kerala (Engg.) 2005]
Options:
A) (2, 1)
B) (1, 2)
C) (1, 2)
D) (-1, 2)
Show Answer
Answer:
Correct Answer: C
Solution:
As the line $ ax+by=0 $ touches the circle $ x^{2}+y^{2}+2x+4y=0 $ , distance of the centre (-1, -2) from the line = radius
therefore $ | \frac{-a-2b}{\sqrt{a^{2}+b^{2}}} |=\sqrt{{{(-1)}^{2}}+{{(-2)}^{2}}} $
therefore $ {{(a+2b)}^{2}}=5(a^{2}+b^{2}) $
therefore $ 4a^{2}-4ab+b^{2}=0 $
therefore $ {{(2a-b)}^{2}}=0 $ \ $ b=2a $ . Next, $ ax+by=0 $ is a normal to $ x^{2}+y^{2}-4x+2y-3=0 $ , the centre (2, -1) should lie on $ ax+by=0 $ \ $ 2a-b=0\Rightarrow b=2a $ .
Hence $ a=1 $ , $ b=2 $ .
BETA
