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Let $x_0, y_0$ be fixed real numbers such that $x_0^2+y_0^2>1$. If $x,y$ are arbitrary real numbers such that $x^2+y^2\leq1$, then the minimum value of the expression $(x-x_0)^2+(y-y_0)^2$ is?

A) $(\sqrt{x_0^2+y_0^2}-1)^2$

B) $x_0^2+y_0^2-1$

C)$(|x_0|+|y_0|-1)^2$

D)$(|x_0|+|y_0|)^2-1$

How do I approach this question?

I see that the required expression $(x-x_0)^2+(y-y_0)^2$ is the equation of a circle with center at $(x_0,y_0)$ and its minimum value would be its radius, maybe?

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The number $(x-x_0)^2+(y-y_0)^2$ is the square of the distance from $(x,y)$ to $(x_0,y_0)$. So, the minimum is attained at the point of the circle $\{(x,y)\in\mathbb R^2\mid x^2+y^2\leqslant 1\}$ which is closest to $(x_0,y_0)$, which is$$\left(\frac{x_0}{\sqrt{{x_0}^2+{y_0}^2}},\frac{y_0}{\sqrt{{x_0}^2+{y_0}^2}}\right).$$Can you take from here?

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  • $\begingroup$ Got it, thanks! Posted how I solved it $\endgroup$ – Techie5879 Oct 16 '19 at 7:21
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Solved it, first I constructed a circle with the equation $x^2+y^2-1 = 0$, thus center at origin. Then I put $x_0$ and $y_0$ in the equation to get $x_0^2+y_0^2-1$, which is greater than $0$ (since $x_0^2+y_0^2 >1$) so it must lie outside the circle. So, required distance would be the distance of $(x_0,y_0)$ from $(0,0)$ and minus the radius, which would be $(\sqrt{x_0^2+y_0^2}-1)$, corresponding to (A)

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