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Grade:

                        

find the value of k in which straight line 3x+4y=k touches the circle x^2+y^2=10x

7 years ago

Answers : (2)

Himanshu Sharma
31 Points
							

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

 

Please rate if it helped

 

7 years ago
ankitesh gupta
63 Points
							

 

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7 years ago
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