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The value of k for which the circles x^2+y^2=1 and x^2+y^2-4kx+8=0 have only two common tangents is (a) (-9/4,9/4) (b) k 9/4 (c)none of these

The value of k for which the circles x^2+y^2=1 and x^2+y^2-4kx+8=0 have only two common tangents is


(a) (-9/4,9/4)     (b)  k<-9/4 & k>9/4     (c)none of these

Grade:

2 Answers

Himanshu Sharma
31 Points
8 years ago

C:none of these 

x^2+y^2=1    centre=(0,0)   radius=1

x^2+y^2-4kx+8=0   centre(2k,0)   radius=√4k^2-8

for only 2 common tangents sum of radius mustbe larger than the distance between their centres(because now there will be only 2 common tangents both equally inclined to the line joining their centres)

hence

 

2k>1+√4k^2-8

 

2k-1>√4k^2-8

 

as right side is always positive hence we can square both sides without changing the signs

(2k-1)^2>4k^2-8

on solving we get k<9/4..

hope it helped..

Suzuki Mitsuba
4 Points
8 years ago

for only 2 common tangents sum of radius must be larger than the distance between their centres(because now there will be only 2 common tangents both equally inclined to the line joining their centres)

hence

 

2k>1+√4k^2-8??????

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