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

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The discussion revolves around finding the minimum value of the expression (a+7)² + (b+2)² under the constraint (a-5)² + (b-7)² = 4. Participants share their attempts and solutions, with one user expressing appreciation for another's perfect solution. The conversation highlights the challenge of solving the problem mentally and the collaborative nature of the forum. The focus remains on the mathematical problem and its resolution. Overall, the thread emphasizes problem-solving strategies in a communal setting.
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Here is this week's POTW:

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

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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.
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}
:nb)
 
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Still discussing with @anemone but I'd like them to be open and communal as you are doing here :smile:
 
$(a+7)^2+(b+2)^2=\sqrt{ab}$
I like Serena said:
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.
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}
:nb)
@I like Serena, your solution is perfect!
 
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Late to the party, but I did not sneak at Serans solution, I started on this earlier today and just got it finished.

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##
 
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anemone said:
Here is this week's POTW:

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

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To do this in one's head:
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##.
 
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