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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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