To solve this problem I adapt the Lagrange multipliers method used here


This way my attempt to solve this problem, goes as follow:

Let $f(x,y)=(x-8)^{2}+(y-6)^{2}$ the squared distance about the point $(8,6)$ and $$g(x,y)=x^{2}+y^{2}-2x-2y-23=0.$$Then $$\nabla f=2(x-8)i + 2(y-6)j$$ and $$\nabla g = 2(x-1)i + 2(y-1)j.$$ So $\nabla f = \lambda \nabla g$, then $$2(x-8)i +2 (y-6)= \lambda[2(x-1)i+2(y-1)j]$$. This way I obtained $$x=\frac{-8+\lambda}{\lambda-1}$$ and $$y=\frac{-6+\lambda}{\lambda-1}$$. In order to obtain the value of $\lambda$ I substitute the $x$ and $y$ obtained in $$x^{2}+y^{2}-2x-2y-23=0$$. But is really hard to get $\lambda$ from this equation. I was wondering if there is an easier method to solve this problems or if not, I was wondering how to get the value of $\lambda$. And how to conclude this problem?? Thanks!


Only for this particular problem you have the curve $g(x,y)=x^{2}+y^{2}-2x-2y-23=0$, which is equation of a circle $(x-1)^2+(y-1)^2 = 25$ whose radius is $5$ and center at $(1,1)$. So you can easily calculate the distance.

The distance between the center and the point is $\sqrt{(8-1)^2 + (6-1)^2} = \sqrt{74}$, so the minimum distance is $\sqrt{74}-5$, maximum distance is $\sqrt{74}+5$

  • $\begingroup$ Wow! Thanks a lot! Didnt see that nice factorization of $g(x,y)$. No I will practise to became better to find those tricky factorizations. Also, I was wondering If you can find a nice factorization for $g(x,y)=x^2+4x-y^2-4y=8$, I thought it was the circle equation at center $(2,-2)$ and radius 4 but the negative terms at $y$ trouble me :( @amitava $\endgroup$
    – Cos
    May 3 '19 at 20:27
  • $\begingroup$ Sorry for responding late, that is equation of a hyperbola, $(x+2)^2-(y+2)^2 = 8$ and it does not look like it has an easy solution like this. For a particular case (using calculus) you may want to see this math.stackexchange.com/questions/921924 $\endgroup$
    – amitava
    May 4 '19 at 5:02
  • $\begingroup$ @Cos, I am not sure whether you got notification for my earlier comment or not. $\endgroup$
    – amitava
    May 4 '19 at 18:37
  • $\begingroup$ Yes, sorry for the late reply. Changing the variable to solve for an origin centered hyperbola helps a lot! Thank a lot! Still I'm struggling with some other Lagrangian problems I'll be asking here these days :9 @amitava $\endgroup$
    – Cos
    May 4 '19 at 18:51

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