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(3,2) is an exterior point of the circle \mathrm{x^2+y^2-2 x+y=0.} The equation of the circle of minimum radius centred on it containing the given circle is \mathrm{(x-3)^2+(y-2)^2=\left(\frac{\sqrt{k}+\sqrt{5}}{2}\right)^2}, where \mathrm{k=}

Option: 1

40


Option: 2

41


Option: 3

42


Option: 4

39


Answers (1)

best_answer

\mathrm{3^2+2^2-6+4>0} so (3,2) is exterior point of the circle

\mathrm{ x^2+y^2-2 x+y=0 }     ......(1)
The radius of required circle

\mathrm{ =\text { distance between centres }(3,2) \text { and }(1,-1 / 2)+\text { radius of circle (1) } }

\mathrm{=\sqrt{(3-1)^2+\left(2+\frac{1}{2}\right)^2}+\sqrt{1+\frac{1}{4}}=\frac{\sqrt{41}+\sqrt{5}}{2}}

\mathrm{\therefore \text { Equation of required circle is }(x-3)^2+(y-2)^2=\left(\frac{\sqrt{41}+\sqrt{5}}{2}\right)^2}

Posted by

Suraj Bhandari

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