hi
let and two reels such that, ,find the minimum value of .
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):
4. I've got the minimum distance
Here's yet another way to do that: 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 , the other the point where it is maximized.
Far from the easiest way! But now that you have it, look for the value of that minimizes the function by differentiating it with respect to and setting that derivative equal to 0. It will eventually reduce to some constant, again telling you that the extremal points lie on a straight line from the origin.
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 , the other the point where it is maximized.