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  • Equation of the circle whose radius is 5 and which touches externally the circle  <img alt="\mathrm{x^{2}+y^{2}-2 x-4 y-20=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

Equation of the circle whose radius is 5 and which touches externally the circle  \mathrm{x^{2}+y^{2}-2 x-4 y-20=0}  at the point \mathrm{(5,5)} is

Option: 1

\mathrm{(x-9)^{2}+(y-6)^{2}=5^{2}}


Option: 2

\mathrm{(x-9)^{2}+(y-8)^{2}=5^{2}}


Option: 3

\mathrm{(x-7)^{2}+(y-3)^{2}=5^{2}}


Option: 4

none of these

 


Answers (1)

best_answer

If \mathrm{(h, k)} be the centre then (5,5) is the mid-point of \mathrm{(h,k)} and (1,2) because the radii of both the circles are 5 each

\therefore \mathrm{h}=9, \mathrm{k}=8

Hence (B) is the correct answer.

Posted by

manish painkra

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