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  • The locus of the centre of a circle which touches externally the circle <img alt="\mathrm{x^2+y^2-6 x-6 y+14=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E

The locus of the centre of a circle which touches externally the circle \mathrm{x^2+y^2-6 x-6 y+14=0} and also touches the y-axis, is given by the equation

Option: 1

\mathrm{x^2-6 x-10 y+14=0}


Option: 2

\mathrm{x^2-10 x-6 y+14=0}


Option: 3

\mathrm{y^2-6 x-10 y+14=0}


Option: 4

\mathrm{y^2-10 x-6 y+14=0}


Answers (1)

best_answer

Let the centre be (h, k), then radius =h

Also, \mathrm{ C_1 C_2=R_1+R_2}

\mathrm{ \begin{aligned} & \Rightarrow \sqrt{(h-3)^2+(k-3)^2}=h+\sqrt{9+9-14} \\ & \Rightarrow(h-3)^2+(k-3)^2=h^2+4+4 h \\ & \Rightarrow k^2-10 h-6 k+14=0 \end{aligned} }

Thus, locus of \mathrm{(h, k)\, \, is \, \, y^2-10 x-6 y+14=0}

Posted by

Ritika Jonwal

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