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

Tried Cauchy-Schwarz and AM - GM, unable to do.

Kuzja
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    Do you know Lagrange multipliers? – vadim123 May 09 '14 at 13:54
  • Unless I'm missing something, this can be done by high school methods. You simply what the point on the circle $(x+5)^2 +(y-12)^2 =14$ that is furtherest from the origin. Drawing a rough sketch, you'll see that it's the point $\sqrt{14}$ units distant from the point $(-5,12)$ along the line that joins the origin to the point $(-5,12).$ – Dave L. Renfro May 09 '14 at 14:01

4 Answers4

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HINT: For this case where the curve is a circle, the value you seek is square of the distance of the centre of the circle from origin minus the radius. (Draw a diagram to see why?) $$x^2+y^2 = (13-\sqrt{14})^2.$$

You can also find the maximum of $x^2+y^2$ using this trick.

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Another way is triangle inequality (essentially think of the triangle between the origin, the centre of the circle and any point on the circle):

$$\sqrt{x^2+y^2} +\sqrt{(x+5)^2+(y-12)^2} \ge \sqrt{(-5)^2+(12)^2} \implies x^2+y^2 \ge \left(13 - \sqrt{14}\right)^2$$

Macavity
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Hint: take a look at the picture below, and all the problem will vanish...

enter image description here

In fact the picture shows the circle of equation $(x+5)^2 +(y-12)^2 =14$, and the line passing trough its centre and the origin. The question asks the minimum length of the segment whose extremities are the origin and a point on the circumference...Thus you should minimize the distance from the point on the circumference from the origin, and this is done by drawing a line passing through the centre of the square and the centre (trivial proof). So you get $$A\equiv \left(5\left(\frac{\sqrt{14}}{13}-1\right);12\left(1-\frac{\sqrt{14}}{13}\right)\right)\Rightarrow \\ \text{length of} \overline{AB} = \sqrt{\left(5\left(\frac{\sqrt{14}}{13}-1\right)\right)^2+\left(12\left(1-\frac{\sqrt{14}}{13}\right)\right)^2}=\\ \sqrt{(\sqrt{14}-13)^2}=\sqrt{14}-13$$ and finally $$\text{minimum of } x^2+y^2=\overline{AB}^2=(\sqrt{14}-13)^2=183-26\sqrt{14}$$

sirfoga
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  • So, I just solve for the intersection point of the circle and line. get the point A's coordinates and that would be the answer of x,y ? –  May 09 '14 at 14:30
  • The minimum of $x^2+y^2$ is the square of the length of $\overline{AB}$... – sirfoga May 09 '14 at 14:32
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Any point satisfying $\displaystyle(x+5)^2 +(y-12)^2 =14$ can be expressed as $\sqrt{14}\cos\phi-5,\sqrt{14}\sin\phi+12$

$\displaystyle x^2 + y^2=14(\cos^2\phi+\sin^2\phi)+2\sqrt{14}(12\sin\phi-5\cos\phi)+5^2+12^2$ $\displaystyle=14+12^2+5^2+2\sqrt{14}(12\sin\phi-5\cos\phi)$

This will attain minimum if $\displaystyle12\sin\phi-5\cos\phi$ is minimum

Now set $12=r\sin\phi,5=r\cos\phi$ to find $\displaystyle12\sin\phi-5\cos\phi=13\sin\left(\phi-\arctan\frac5{12}\right)$

What is the minimum value of $\displaystyle\sin\left(\phi-\arctan\frac5{12}\right)?$

lab bhattacharjee
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  • @Adeetya, How about this? – lab bhattacharjee May 09 '14 at 18:12
  • for finding minimum value of 12 sin(phi) - 5 cos(phi).. simply use cauchy schwartz inequality... ((12^2) + (-5)^2)) * (sin^2(phi) +cos^2(phi)>=(12sin(phi) -5cos(phi)^2

    ur answer will be 13.

     –  May 10 '14 at 02:26
  • @Adeetya, http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity will put us in the same conclusion, right? The minimum value of $$12\sin\phi-5\cos\phi$$ will be $$-13$$ – lab bhattacharjee May 11 '14 at 04:13