Minimum Value of (a+7)^2+(b+2)^2 with Constraint (a-5)^2+(b-7)^2=4 - POTW #506

[anemone]

Here is this week's POTW:

-----

Find the minimum value of $(a+7)^2+(b+2)^2$ subject to the constraint $(a-5)^2+(b-7)^2=4$.

-----

 

[I like Serena]

I think no procedures or forms are in place, and it's not clear to me if they will be.
In the meantime, here's my attempt.

Spoiler

Geometrically this is the square of the distance of point (-7,-2) to the circle with radius 2 around (5,7).
In turn that is the square of the distance of (-7,-2) to (5,7) reduced by 2.
So the requested minimum value is:
\begin{array}{ll}\Big[d\big((-7,-2), (5,7)\big)-2\Big]^2&=\Big[\sqrt{(5 - -7)^2 + (7 - -2)^2} -2\Big]^2\\
&=\Big[\sqrt{12^2+9^2}-2\Big]^2\\
&=\Big[\sqrt{3^2(4^2+3^2)}-2\Big]^2\\
&=\Big[3\cdot 5 -2\Big]^2\\
&=13^2=169\end{array}

 

[Greg Bernhardt]

Still discussing with @anemone but I'd like them to be open and communal as you are doing here

 

[anemone]

$(a+7)^2+(b+2)^2=\sqrt{ab}$

I think no procedures or forms are in place, and it's not clear to me if they will be.
In the meantime, here's my attempt.

Spoiler

Geometrically this is the square of the distance of point (-7,-2) to the circle with radius 2 around (5,7).
In turn that is the square of the distance of (-7,-2) to (5,7) reduced by 2.
So the requested minimum value is:
\begin{array}{ll}\Big[d\big((-7,-2), (5,7)\big)-2\Big]^2&=\Big[\sqrt{(5 - -7)^2 + (7 - -2)^2} -2\Big]^2\\
&=\Big[\sqrt{12^2+9^2}-2\Big]^2\\
&=\Big[\sqrt{3^2(4^2+3^2)}-2\Big]^2\\
&=\Big[3\cdot 5 -2\Big]^2\\
&=13^2=169\end{array}

@I like Serena, your solution is perfect!

 

[malawi_glenn]

Late to the party, but I did not sneak at Serans solution, I started on this earlier today and just got it finished.

Spoiler

There is probably a nice geometric way of solving this in five lines. I did brute force method with calculus and quadratic equations.
$(a-5)^2 + (b-7)^2=4$ solve for $b$:
$b_+ = 7 + \sqrt{10a-a^2-21}$
$b_- = 7 - \sqrt{10a-a^2-21}$
Construct functions:
$f_+(a)= (a+7)^2+(b_++2)^2 = 24a + 18\sqrt{10a-a^2-21} + 109$
$f_-(a)= (a+7)^2+(b_-+2)^2 = 24a - 18\sqrt{10a-a^2-21} + 109$
Domain of both funcions is $3 \leq a \leq 7$ and we have $f_+(3) = f_-(3) = 181$ and $f_+(7) = f_-(7) = 277$
Differentiate w.r.t. $a$ and solve derivative equal to zero.
$\dfrac{\mathrm{d} f_+ }{\mathrm{d} a }= \dfrac{90-18a}{\sqrt{10a-a^2-21}}+ 24$
$\dfrac{\mathrm{d} f_+ }{\mathrm{d} a }= 0$ we can solve $90-18a = 24\sqrt{10a-a^2-21}$ by squaring both sides and look out for false roots, though one of them will be the root for $\dfrac{\mathrm{d} f_- }{\mathrm{d} a }= 0$, see below. The root is $a = \dfrac{33}{5}$
$f_+(\frac{33}{5}) = 289$
Now the other function
$\dfrac{\mathrm{d} f_- }{\mathrm{d} a }= \dfrac{18a-90}{\sqrt{10a-a^2-21}}+ 24$
$\dfrac{\mathrm{d} f_- }{\mathrm{d} a }= 0$ has root $a = \dfrac{17}{5}$
$f_-(\frac{17}{5}) = 169$

Smallest value $169 = 13^2$
Largest value $289 = 17^2$

 

[PeroK]

Here is this week's POTW:

-----

Find the minimum value of $(a+7)^2+(b+2)^2$ subject to the constraint $(a-5)^2+(b-7)^2=4$.

-----

To do this in one's head:

Spoiler

Let $x = a -5, y = b - 7$, then we have the minimim value of $(x + 12)^2 + (y + 9)^2$ subject to $(x, y)$ lying on the circle at the origin of radius $2$. $(12, 9)$ forms a Pythagorean triple with $15$, so the minimum distance is $15 - 2 = 13$. The square of which is $169$.