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  • The equation of the circle of radius which touches the line <img alt="x+y=1 at \: (2,-1)" src="https://en

The equation of the circle of radius \sqrt{2} which touches the line x+y=1 at \: (2,-1) is

Option: 1

\mathrm{x^{2}+y^{2}-4 x+2 y+3=0}


Option: 2

\mathrm{x^{2}+y^{2}+6 x+7=0}


Option: 3

\mathrm{x^{2}+y^{2}-2 x+4 y+3=0}


Option: 4

none of these


Answers (1)

best_answer

Family of circle touches \mathrm{x+y-1=0} at \mathrm{(2,-1)} is

\mathrm{(x-2)^{2}+(y+1)^{2}+\lambda(x+y-1)=0}

Now radius of this circle is \mathrm{\sqrt{2}}

Hence \mathrm{I= \pm 2}

Hence (C) is the correct answer.

Posted by

chirag

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