# Thread: minimum value

1. ## minimum value

hi
let $\displaystyle x$ and $\displaystyle y$ two reels such that,$\displaystyle (x+5)^2+(y-12)^2=14^2$,find the minimum value of $\displaystyle x^2+y^2$.
thanks.

2. Originally Posted by Raoh
hi
let $\displaystyle x$ and $\displaystyle y$ two reels such that,$\displaystyle (x+5)^2+(y-12)^2=14^2$,find the minimum value of $\displaystyle x^2+y^2$.
thanks.
1. The first equation describes a circle around M(-5, 12) with r = 14.
The 2nd term belongs to a circle around the origin with a variable radius.

2. The minimum of x² + y² = r² occurs when the 2nd circle is tangent to the first circle. In this case both midpoints and the tangent point are placed on a straight line through the origin (= center of the 2nd circle)

3. Determine the tangent point by solving for (x, y):

$\displaystyle \left|\begin{array}{rcl}(x+5)^2+(y-12)^2&=&14^2 \\ y &=& -\frac{12}5 x \end{array} \right.$

4. I've got the minimum distance $\displaystyle d_{min} = 1$

3. thank you.
i have another approach i hope you would help me finish it.
let $\displaystyle x+5=14\cos(\phi)$ and $\displaystyle y-12=14\sin(\phi)$ such that $\displaystyle \phi\in [0,2\pi )$
i ended up with,
$\displaystyle x^2+y^2=\left (14\cos(\phi )-5 \right )^2+(12+14\sin(\phi))^2$$\displaystyle =196 \cos^2(\phi)-140 \cos(\phi)+25+196 \sin^2(\phi)+336 \sin(\phi)+144=$$\displaystyle 196-140 \cos(\phi)+336 \sin(\phi)+169=365-140 \cos(\phi)+336 \sin(\phi)$
any ideas how to proceed ?
thanks.

4. Here's yet another way to do that: $\displaystyle x^2+ y^2$ is the square of the distance from (0, 0) to (x, y) and will be minimized when the distance is- and the shortest distance is always on a straight line. Draw a Straight line from (0, 0) to (-5, 12), center of the circle. The point on the circle closest to (0, 0) will be the point where that line crosses the circle. Find the equation of that line (since it passes through (0, 0) it can be written as y= mx) and solve that and the equation of the circle. That will result in a quadratic which has two solutions- one solution will be point that minimizes $\displaystyle x^2+ y^2$, the other the point where it is maximized.

5. Originally Posted by Raoh
thank you.
i have another approach i hope you would help me finish it.
let $\displaystyle x+5=14\cos(\phi)$ and $\displaystyle y-12=14\sin(\phi)$ such that $\displaystyle \phi\in [0,2\pi )$
i ended up with,
$\displaystyle x^2+y^2=\left (14\cos(\phi )-5 \right )^2+(12+14\sin(\phi))^2$$\displaystyle =196 \cos^2(\phi)-140 \cos(\phi)+25+196 \sin^2(\phi)+336 \sin(\phi)+144=$$\displaystyle 196-140 \cos(\phi)+336 \sin(\phi)+169=365-140 \cos(\phi)+336 \sin(\phi)$
any ideas how to proceed ?
thanks.
Far from the easiest way! But now that you have it, look for the value of $\displaystyle \phi$ that minimizes the function by differentiating it with respect to $\displaystyle \phi$ and setting that derivative equal to 0. It will eventually reduce to $\displaystyle tan(\phi)=$ some constant, again telling you that the extremal points lie on a straight line from the origin.

6. Originally Posted by HallsofIvy
Far from the easiest way! But now that you have it, look for the value of $\displaystyle \phi$ that minimizes the function by differentiating it with respect to $\displaystyle \phi$ and setting that derivative equal to 0. It will eventually reduce to $\displaystyle tan(\phi)=$ some constant, again telling you that the extremal points lie on a straight line from the origin.
thank you "Hallsofivy"

7. Originally Posted by HallsofIvy
Here's yet another way to do that: $\displaystyle x^2+ y^2$ is the square of the distance from (0, 0) to (x, y) and will be minimized when the distance is ..?.. and the shortest distance is always on a straight line. Draw a Straight line from (0, 0) to (-5, 12), center of the circle. The point on the circle closest to (0, 0) will be the point where that line crosses the circle. Find the equation of that line (since it passes through (0, 0) it can be written as y= mx) and solve that and the equation of the circle. That will result in a quadratic which has two solutions- one solution will be point that minimizes $\displaystyle x^2+ y^2$, the other the point where it is maximized.
thanks but i think there's something missing.

8. Originally Posted by Raoh
thank you.
i have another approach i hope you would help me finish it.
let $\displaystyle x+5=14\cos(\phi)$ and $\displaystyle y-12=14\sin(\phi)$ such that $\displaystyle \phi\in [0,2\pi )$
i ended up with,
$\displaystyle x^2+y^2=\left (14\cos(\phi )-5 \right )^2+(12+14\sin(\phi))^2$$\displaystyle =196 \cos^2(\phi)-140 \cos(\phi)+25+196 \sin^2(\phi)+336 \sin(\phi)+144=$$\displaystyle 196-140 \cos(\phi)+336 \sin(\phi)+169=365-140 \cos(\phi)+336 \sin(\phi)$
any ideas how to proceed ?
thanks.
1. All your calcualtions are correct!

2. You have now a function in $\displaystyle \phi$ of which you want to get a minimum: Let $\displaystyle s(\phi)$ denote the function:

$\displaystyle s(\phi)=365-140 \cos(\phi)+336 \sin(\phi)$

Differentiate wrt $\displaystyle \phi$ and solve the equation $\displaystyle s'(\phi) = 0$ for $\displaystyle \phi$

3.
$\displaystyle 336 \cos(\phi) + 140 \sin(\phi) = 0$

4. Divide the equation by $\displaystyle \cos(\phi)$ and keep in mind that $\displaystyle \frac{\sin(\phi)}{\cos(\phi)} = \tan(\phi)$

5. Plug in the result of $\displaystyle \phi$ into $\displaystyle s(\phi)$.

9. Originally Posted by HallsofIvy
Here's yet another way to do that: $\displaystyle x^2+ y^2$ is the square of the distance from (0, 0) to (x, y) and will be minimized when the distance is
No, that's perfectly correct English: "The square of the distance will be minimized when the distance is". That's equivalent to "The square of the distance will be minimized when the distance is minimized".

- and the shortest distance is always on a straight line. Draw a Straight line from (0, 0) to (-5, 12), center of the circle. The point on the circle closest to (0, 0) will be the point where that line crosses the circle. Find the equation of that line (since it passes through (0, 0) it can be written as y= mx) and solve that and the equation of the circle. That will result in a quadratic which has two solutions- one solution will be point that minimizes $\displaystyle x^2+ y^2$, the other the point where it is maximized.
Originally Posted by Raoh
thanks but i think there's something missing.