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  • Equation of a circle with centre touching the circle <img alt="\mathrm{x^{2}+y^{2}=1}" src="https://entrancecor

Equation of a circle with centre (4,3) touching the circle \mathrm{x^{2}+y^{2}=1} is

Option: 1

\mathrm{x^{2}+y^{2}-8 x-6 y-9=0}


Option: 2

\mathrm{x^{2}+y^{2}-8 x-6 y+11=0}


Option: 3

\mathrm{x^{2}+y^{2}-8 x-6 y+11=0}


Option: 4

\mathrm{x^{2}+y^{2}-8 x-6 y+9=0}


Answers (1)

best_answer

Let the circle touching the circle \mathrm{x^{2}+y^{2}=1, be \: x^{2}+y^{2}-8 x-6 y+k=0},
The equation of the common tangent is \mathrm{S_{1}-S_{2}=0}

i.e. \mathrm{8 x+6 y-1-k=0}
This is a tangent to the circle \mathrm{x^{2}+y^{2}=1}.

Hence \mathrm{\pm 1=\frac{k+1}{\sqrt{8^{2}+6^{2}}}}

\mathrm{\Rightarrow \mathrm{k}+1= \pm 10 \Rightarrow \mathrm{k}=-11} or 9

Therefore the circles are \mathrm{x^{2}+y^{2}-8 x-6 y+9=0\: and \: x^{2}+y^{2}-8 x-6 y-11=0}

 Hence (D) is the correct answer.

Posted by

shivangi.bhatnagar

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