JEE Main On 16 April 2018 Question 21
Question: If a circle C, whose radius is 3, touches externally the circle, $ x^{2}+y^{2}+2x-4y-4=0 $ at the point $ (2,2) $ , then the length of the intercept cut by this circle C, on the x-axis is equal to. [JEE Main 16-4-2018]
Options:
A) $ \sqrt{5} $
B) $ 2\sqrt{3} $
C) $ 3\sqrt{2} $
D) $ 2\sqrt{5} $
Show Answer
Answer:
Correct Answer: D
Solution:
- Center of $ x^{2}+y^{2}+2x-4y-4=0 $ is at $ (-1,2) $ and radius is Let $ A=(x,y) $ be the center of the circle C, then $ x-1=4\Rightarrow x=5 $ and $ y+2=4\Rightarrow y=2. $ So the center of C is (5, 2) and its radius is 3 The length of the intercept it cuts on the x-axis is $ 2\sqrt{9-4}=2\sqrt{5}. $ So option D is the correct answer. $ x^{2}+y^{2}+2x-4y-4=0 $ Centre $ =(-1,2) $ radius $ =\sqrt{1^{2}+2^{2}+4} $ $ =\sqrt{9} $ = 3 $ ( \frac{h-1}{2},\frac{k+2}{2} )=(2,2) $ $ h-1=4, $ and $ k+2=8 $
$ \Rightarrow $ $ h=5,k=2 $ $ C:{{(x-5)}^{2}}+{{(y-2)}^{2}}=3^{2} $ $ p=\frac{|2|}{\sqrt{1}}=2 $ $ AM=\sqrt{3^{2}-2^{3}}=\sqrt{5} $ $ AB=2AB=2\sqrt{5}. $
