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The radius of the circle passing through the point (6,2), two of whose diameters are \mathrm{x+y=6} and \mathrm{x+2 \mathrm{y}=4} is

Option: 1

10


Option: 2

2 \sqrt{5}


Option: 3

6


Option: 4

4


Answers (1)

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The two diameters intersect at (8,-2) which is the centre of the circle. The circle passes through (6, 2),
 therefore its radius =\sqrt{(8-6)^{2}+(-2-2)^{2}}=\sqrt{20}. Hence the equation of the circle is (x-8)^{2}+(y$ $+2)^{2}=(\sqrt{20})^{2}.

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Ritika Kankaria

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