Let x and y be real numbers satisfying the eq | vedclass

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(B) The given equation is $x^2 - 4x + y^2 + 3 = 0$.Completing the square for $x$,we get $(x^2 - 4x + 4) + y^2 = 4 - 3$,which simplifies to $(x - 2)^2

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$8$

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(B) The given equation is $x^2 - 4x + y^2 + 3 = 0$.Completing the square for $x$,we get $(x^2 - 4x + 4) + y^2 = 4 - 3$,which simplifies to $(x - 2)^2 + y^2 = 1$.This represents a circle with center $(2, 0)$ and radius $r = 1$.Let $x^2 + y^2 = k$,which represents the square of the distance from the origin $(0, 0)$ to any point $(x, y)$ on the circle.The distance from the origin to the center $(2, 0)$ is $d = \sqrt{(2-0)^2 + (0-0)^2} = 2$.The maximum distance from the origin to a point on the circle is $d + r = 2 + 1 = 3$,so $M = 3^2 = 9$.The minimum distance from the origin to a point on the circle is $d - r = 2 - 1 = 1$,so $m = 1^2 = 1$.Thus,$M - m = 9 - 1 = 8$.

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

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