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Total number of common tangents of \mathrm{x^2+y^2-2 x-4 y=0}  and \mathrm{x^2+y^2-8 y-4=0}, is equal to
 

Option: 1

1


Option: 2

2


Option: 3

4


Option: 4

none of these


Answers (1)

best_answer

Centre and radius of \mathrm{x^2+y^2-2 x-4 y=0}  is (1,2) and \sqrt{5}.
Similarly centre and radius of \mathrm{x^2+y^2-8 y-4=0}  is (0,4) and \sqrt{20}

Distance between centres = \sqrt{1+4}=\sqrt{5}

Which is same as difference between radii. Thus circles touch each other internally and hence there will only be one common tangents.

Posted by

Ritika Kankaria

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