Let C be a circle, and let P be a point not on the circle. Prove that the maximum and minimum distances from P to a point X on C occur when the line X P goes through the center of C. [Hint: Choose coordinate systems so that C is defined by x2 + y2 = r2 and P is a point (a,0) on the x-axis with a neq pm r, use calculus to find the maximum and minimum for the square of the distance. Don’t forget to pay attention to endpoints and places where a derivative might not exist.]

Question
Alternate coordinate systems
Let C be a circle, and let P be a point not on the circle. Prove that the maximum and minimum distances from P to a point X on C occur when the line X P goes through the center of C. [Hint: Choose coordinate systems so that C is defined by
$$x2 + y2 = r2$$ and P is a point (a,0)
on the x-axis with a $$\neq \pm r,$$ use calculus to find the maximum and minimum for the square of the distance. Don’t forget to pay attention to endpoints and places where a derivative might not exist.]

2020-12-15
The equation for C is $$x^{2} + y^{2} = r^{2}$$ and P is a point (a, 0)
on the x-axis with a $$\neq \pm r.$$
The point on the circle C is(x, y).
Obtain the distance between (x, y) and (a, 0) as follows:
$$d =\sqrt{(x - a)^2 + (y - 0)^2}$$
$$=\sqrt{(x-a)^{2}+y^{2}}$$
Now, obtain the partial derivatives as follows,
$$d_{x} = 0$$
$$\frac{2(x-a)}{2\sqrt{(x-a)^{2}+y^{2}}}=0$$
$$x = a$$
$$d_{y} = 0$$
$$\frac{2y}{2\sqrt{(x-a)^{2}+y^{2}}}=0$$
$$y = 0$$
Thus, the critical point is(a, 0). However, this point doesn’t lie on the circle C.
We know the endpoints on the circles are (-r, 0) and (r, 0).
Thus, the distance becomes
At (-r, 0),
$$d = \sqrt{(-r-a)^{2}+(0)^{2}}$$
$$=\sqrt{(r+a)^{2}}$$
$$= r + a$$
and at (r, 0),
$$d=\sqrt{(-r-a)^{2}+(0)^{2}}$$
$$=\sqrt{(r-a)^{2}}$$
$$= r - a$$
Apart from these two points, no other point gives these distances but between $$r + a, r - a.$$
Also, these two points lie on the line from (a, 0) that passes through the center of C.
Thus, the maximum distance from the point P (a, 0) to the point $$X (x, y) on C is r + a$$
and the minimum distance from the point P (a, 0) to the point $$X (x, y) on C is r - a.$$
Hence, the maximum and minimum distances from P toa point X on C occur when the line XP goes through the center of C.

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