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find the value of k in which straight line 3x+4y=k touches the circle x^2+y^2=10x
it''s quite simple
simply first put the eqn of line in the curve
3x+4y=k
(k-3x)/4=y
put this in the curve
x^2+(k-3x/4)^2=10x
16x^2+k^2+9x^2-6kx=160x
25x^2+x(-6k-160)+k^2=0
now this quadratic equation will have only one solution because this line touches it at only one point hence the value of x should come out only one.
so Discriminant of this eqn is 0(for same roots)
36k^2+25600+1920k-4*25*k^2=0
hence k=40,-10
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