MHB Minimize r/(sqrt(2)) Distance from (2r, 0) & (0, 2r)

[leprofece]

find a point above the circle: x² + y² = r² so that the sum of the squares of their distances to points (2r, 0) and (0, 2r) it is the smallest

answer (r/(sqrt(2)), r/(sqrt(2)) )

according to the problem
after elevating to the square
(x-2r)²+y² +x²+(y-2r)²

Solving
2(x)²-4rx-8ry+2y²

now y²= r-x²

introducing in one and derivating i don't get the answer
because 8 ry must be 8r( sqrt r-x²)

 

[Evgeny.Makarov]

find a point above the circle: x² + y² = r² so that the sum of the squares of their distances to points (2r, 0) and (0, 2r) it is the smallest

answer (r/(sqrt(2)), r/(sqrt(2)) )

Why not $(r,r)$?

after elevating to the square
(x-2r)²+y² +x²+(y-2r)²

Solving
2(x)²-4rx-8ry+2y²

Why is this expression not symmetric w.r.t. $x$ and $y$?

 

[leprofece]

Why not $(r,r)$?

Why is this expression not symmetric w.r.t. $x$ and $y$?

I don't know this is my thinking maybe it is easier for you another way

 

[Evgeny.Makarov]

You wrote, "Find a point above the circle", but the supposed answer is a point on the circle. Since the segment connecting $(2r,0)$ and $(0,2r)$ does not intersect the circle, the middle of this segment is the point that minimizes the required expression.

 

[leprofece]

You wrote, "Find a point above the circle", but the supposed answer is a point on the circle. Since the segment connecting $(2r,0)$ and $(0,2r)$ does not intersect the circle, the middle of this segment is the point that minimizes the required expression.

yEAH IT IS on NOT ABOVE

 

[Evgeny.Makarov]

after elevating to the square
(x-2r)²+y² +x²+(y-2r)²

Solving
2(x)²-4rx-8ry+2y²

This expression should be
\[
2x^2+2y^2-4r(x+y)
\]
As I said, it is symmetric with respect to $x$ and $y$. Since $x^2+y^2=r^2$, it can be rewritten as $2r^2-4r(x+y)$. Minimizing it is equivalent to maximizing $x+y$. One thing that comes to mind is representing this as $\cos\varphi+\sin\varphi$ where $0\le\varphi<2\pi$. Differentiating this with respect to $\varphi$ and equating to 0, we get $-\sin\varphi+\cos\varphi=0$, or $\tan\varphi=1$, which means $\varphi=\pi/4$ or $\varphi=5\pi/4$. In both cases, $x=y$.

 

[leprofece]

This expression should be
\[
2x^2+2y^2-4r(x+y)
\]
As I said, it is symmetric with respect to $x$ and $y$. Since $x^2+y^2=r^2$, it can be rewritten as $2r^2-4r(x+y)$. Minimizing it is equivalent to maximizing $x+y$. One thing that comes to mind is representing this as $\cos\varphi+\sin\varphi$ where $0\le\varphi<2\pi$. Differentiating this with respect to $\varphi$ and equating to 0, we get $-\sin\varphi+\cos\varphi=0$, or $\tan\varphi=1$, which means $\varphi=\pi/4$ or $\varphi=5\pi/4$. In both cases, $x=y$.

in my book appears I´d really appreciatte that
or I really appreciatte that then