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  • If <img alt="\mathrm{x^2+y^2+4 x+2 y=0 \: and \: x^2+y^2+2 x+k y=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;4%20x&plus;2%20y%3D0%20%5C%3A

If \mathrm{x^2+y^2+4 x+2 y=0 \: and \: x^2+y^2+2 x+k y=0} touch each other, then \mathrm{k} is equal to
 

Option: 1

4
 


Option: 2

1


Option: 3

\frac{1}{2}


Option: 4

2


Answers (1)

best_answer

Both circles pass through origin, so they must touch at origin and will have common tangent at (0,0).
Equation of tangent for Ist circle at (0,0) is

\mathrm{ 2 x+y=0 }..........(i)

and for 2^{\text {nd }} circle is \mathrm{2 x+k y=0}..........(ii)

(i) and (ii) are equation of common tangent.

\therefore Compare (i) and (ii), \mathrm{k=1}

Hence option 2 is correct.

Posted by

mansi

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