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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
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..
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..
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??????
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)
2k>1+√4k^2-8??????
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