Question 1209658
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Answer: <font color=red>3</font>


Explanation


If you complete the square for the x and y terms, then you'll go from
{{{x^2 + y^2 - 8x + 6y + 23 = 0}}}
to
{{{(x-4)^2+(y+3)^2 = 2}}}
This is a circle with center (4,-3). The radius is {{{r = sqrt(2)}}}
You can use various tools such as GeoGebra or WolframAlpha to verify the claim. 
Or you can expand out the terms in the 2nd equation to arrive back at the 1st equation.


This will mean the point (x,y) is on the circle's boundary.


Consider the equation x+y = k
The goal is to find the largest k value possible.


If you draw out the graph of the circle, and draw various lines of the form x+y = k, then you should notice that exactly one line will be tangent to the circle at the northeast corner.
{{{
drawing(400,400,-2,6,-6+1.5,2+1.5,
graph(400,400,-2,6,-6+1.5,2+1.5,-100,-x+3),
circle(4,-3,sqrt(2)),
circle(4,-3,0.05),circle(4,-3,0.07),
circle(5,-2,0.05),circle(5,-2,0.07),
locate(4+0.1,-3-0.1,"A"),locate(5+0.1,-2+0.2,"B"),locate(1.5,2,"x+y=3")
)
}}}
A = center of the circle = (4,-3)
B = tangent point = (5,-2)
The green tangent line has the equation x+y = <font color=red>3</font>
This is the largest x+y can get when subjected to the condition that {{{x^2 + y^2 - 8x + 6y + 23 = 0}}} aka {{{(x-4)^2+(y+3)^2 = 2}}}
Note radius AB has slope 1 which is the negative reciprocal of the tangent slope -1.
Radius AB is perpendicular to the green tangent line.


Check out this interactive <a href="https://www.desmos.com/calculator/ke5vswmyda">Desmos graph</a>
Adjusting the slider for parameter k will move the line x+y = k up or down.


Verification using <a href="https://www.wolframalpha.com/input?i=max+x%2By+when+x%5E2+%2B+y%5E2+-+8x+%2B+6y+%2B+23+%3D+0">WolframAlpha</a>
The answer on the WolframAlpha page is a bit buried in a sea of numbers & symbols, but it does mention <font color=red>3</font> under the "global maximum" subsection and just before the "at (x,y) = (5,-2)"


More practice with a similar problem is found <a href="https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.1208800.html">here</a>
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