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  • The point of tangency of the circles and <img alt="x^{2}+y^{2}-8y-

The point of tangency of the circles x^{2}+y^{2}-2x-4y=0 and x^{2}+y^{2}-8y-4=0, is

Option: 1

\left ( 2,0 \right )


Option: 2

\left ( 0,2 \right )


Option: 3

\left ( 4,0 \right )

 


Option: 4

\left ( 0,-4 \right )


Answers (1)

best_answer

x^{2}+y^{2}-2x-4y=0\rightarrow \text {Centre}\; C_{1}\left ( 1,2 \right ), R_{1}=\sqrt{5}

x^{2}+y^{2}-8y-4=0\rightarrow \text {Centre}\; C_{2}\left ( 0,4 \right ), R_{2}=2\sqrt{5}

C_{1}C_{2}=\sqrt{5}=R_{1}\sim R_{2'}

\therefore Circles internally touch each other at the point \left ( 2,0 \right )

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mansi

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