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  • The locus of the centre of a circle of radius 2 which rolls outside on the circumference of circle <img alt="\mathrm{x^2+y^2+3 x-6 y-9=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cma

The locus of the centre of a circle of radius 2 which rolls outside on the circumference of circle \mathrm{x^2+y^2+3 x-6 y-9=0} is

 

Option: 1

\mathrm{x^2+y^2+3 x-6 y+5=0}



 


Option: 2

\mathrm{x^2+y^2+3 x-6 y-31=0 }


Option: 3

\mathrm{ x^2+y^2+3 x-6 y+\frac{19}{4}=0 }


Option: 4

none of these


Answers (1)

best_answer

If centre of the circle which rolls outside the given circle be \mathrm{(h, k)}, then distance between the centres = sum of radii.
\mathrm{ \Rightarrow \quad(h+3 / 2)^2+(k-3)^2=\left(\frac{9}{2}-2\right)^2 \Rightarrow h^2+k^2+3 h-6 k-31=0 }

Hence option 2 is correct.

Posted by

Sanket Gandhi

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