question_answer

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]

A) \[\sqrt{5}\]                              

B) \[2\sqrt{3}\]

C)  \[3\sqrt{2}\]               

D)  \[2\sqrt{5}\]

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 \[\left( \frac{h-1}{2},\frac{k+2}{2} \right)=(2,2)\] \[h-1=4,\]and \[k+2=8\] \[\Rightarrow \]\[h=5,k=2\] \[C:{{(x-5)}^{2}}+{{(y-2)}^{2}}={{3}^{2}}\] \[p=\frac{|2|}{\sqrt{1}}=2\] \[AM=\sqrt{{{3}^{2}}-{{2}^{3}}}=\sqrt{5}\] \[AB=2AB=2\sqrt{5}.\]