Question

# Let x & y be the real numbers satisfying the equation $$x^2-4x+y^2+3=0$$.If the maximum and minimum values of $$x^2+y^2$$ are M & m respectively , then find the numerical value of (M+m).

Solution

## $$x^2-4x+y^2+3=0$$$$\Rightarrow x^2-4x+4+y^2-1=0$$    $$\left( x-2\right)^2+y^2=1$$represents a circle with center at $$\left( 2,0\right)$$ $$\&$$ radius $$=1$$ units$$\therefore x^2+y^2\le 3$$    $$x^2+y^2\ge 1$$$$\therefore M=3,m=1$$$$\therefore M+m=4$$Hence, the answer is $$4.$$Maths

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