# How do I find the minimum value of the following expression?

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?

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?
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)