Minimum value of $x^2+y^2$

Question

The problem is as follows:

Find the values of $x,y$ for which $x^2+y^2$ takes the minimum value where $(x+5)^2+(y-12)^2=14$.

I was trying to solve it using the extremal principle. But, I couldn't get any far. Can anyone give me some hints?

Answer 1

The problem is asking for the point on the given circle closest to the origin. Draw a line through the origin and $(-5,12)$; it will intersect the circle at a distance of $\sqrt{(-5)^2 + 12^2} - \sqrt{14} = 13-\sqrt{14}$ from the origin.

The coördinates of the point we are looking for will therefore be:

$$(x,y) = \frac{13-\sqrt{14}}{13}(-5,12).$$

Answer 2

Solution without using Lagrangian:

Let $$x=-5+\sqrt{14}\cos t$$ $$y=12+\sqrt{14}\sin t$$ \begin{align} x^2+y^2&=5^2-10\sqrt{14}\cos t+14\cos^2 t+12^2+24\sqrt{14}\sin t+14\sin^2 t\\ &=5^2+12^2+14(\cos^2 t + \sin^2 t)-2\sqrt{14}(5\cos t-12\sin t)\\ &=5^2+12^2+14-2\sqrt{14}\sqrt{5^2+12^2}\cos(t+\phi) \end{align} As $-1\le\cos(t+\phi)\le1$, $$183-26\sqrt{14}\le x^2+y^2\le 183+26\sqrt{14}$$

Answer 3

Using Lagrange multipliers, consider $$F=x^2+y^2+\lambda\big((x+5)^2+(y-12)^2-14\big)$$ which gives $$F'_x=2 \lambda (x+5)+2 x$$ $$F'_y=2 \lambda (y-12)+2 y$$ $$F'_\lambda=(x+5)^2+(y-12)^2-14$$ from which $$F'_x=0\implies x=-\frac{5 \lambda }{\lambda +1}$$ $$F'_y=0\implies y=\frac{12 \lambda }{\lambda +1}$$ Plugging in the last partial derivative $$F'_\lambda=0 \implies \frac{169}{(\lambda +1)^2}-14=0$$ from which two solutions $$\lambda_1=-1-\frac{13}{\sqrt{14}}$$ $$\lambda_2=-1+\frac{13}{\sqrt{14}}$$ Computing the values of $x$ and $y$ for $\lambda_1$ yields to $x^2+y^2=183+26 \sqrt{14}$; doing the same for $\lambda_2$ yields to $x^2+y^2=183-26 \sqrt{14}$.

This is the same as what Kay K. elegantly answered.