Minimum point of $x^2+y^2$ given that $x+y=10$

Since the condition $x+y=10$ is simple and allows easy elemination of one variable, one possible approach is to put the resulting $y=10-x$ into the term to mimimize:

$$x^2+y^2=x^2+(10-x)^2:=f(x)$$

You now have a function in one variable (named $f(x)$) where you are looking for the minimal value over all real $x$.

Since this is a quadratice function, it can be minimized with a bit of algebra and without calculus:

$$f(x)=x^2+(10-x)^2=2x^2-20x+100=2(x^2-10x)+100 = 2(x^2-10x+25) + 50 = 2(x-5)^2+50 \ge 50.$$

So we have $f(x)\ge 50$ and equality happens at $x=5$ (which implies $y=5$).

Since the tag "derivatives" is used, I'll also use the usual calculus approach:

We have $f(x) = 2x^2-20x+100$, which implies $f'(x)=4x-20$ and $f''(x)=4$.

A local mimimum $x_m$ has $f'(x_m)=0$ as necessay condition, and $4x_m-20=0$ easily leads to the only solution $x_m=5$ and $f''(x_m)=4 > 0$ shows this is a local mimimum, with $f(x_m)=f(5)=50$.

Also, one needs to check the behaviour of $f(x)$ when $x$ tends to $+\infty$ and $-\infty$, as $f'(x_m)=0$ only finds local extrema. Since $f(x)$ is a quadratic with positive constant before $x^2$, the function tends to $+\infty$ in either case, so no interference with the looked for minimum.

Take the vectors $u=(1,1)$ and $v=(x,y)$ then apply Cauchy-Schwarz inequality. It comes : $2(x^2+y^2)\geq (x+y)^2=100$ and for $x=y=5$ the equality holds. The minimum value is therefore $50$.

Showing the problem graphically

The line is defined by the equation $f: x + y = 10$. Minimizing $x^2 + y^2$ amounts to finding the smallest circle around the origin touching the line. Since all radii are perpendicular, we search for the intersection of $f$ and $y - x = 0$ which is solved by $B: x = y = 5$.