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Radius of the smallest circle passing through the point of intersection of circles \mathrm{x^{2}+y^{2}+2 x-3} and \mathrm{x^{2}+y^{2}+3 x-y=0} is

Option: 1

1


Option: 2

\sqrt{2}


Option: 3

\sqrt{2}


Option: 4

None of these.


Answers (1)

best_answer

\mathrm{x^{2}+y^{2}+2 x-3=0}\quad \ldots(1)
\mathrm{x^{2}+y^{2}+3 x-y=0}\quad \ldots(2)

Equation of the common chord is
\mathrm{x-y+3=0}\quad \ldots(3)
Perpendicular distance of  \mathrm{P(-1,0)} centre of circle (1), from the common chord (3) is \mathrm{\frac{-1+3}{\sqrt{2}}=\sqrt{2}}
and radius of the circle (1) is \mathrm{\sqrt{1+3}=2}

Radius of the smallest circle is \mathrm{\sqrt{2^{2}-(\sqrt{2})^{2}}=\sqrt{2}}.

 Hence (B) is the correct answer.

Posted by

avinash.dongre

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