# 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

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 $\ne ±r,$ 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.]
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Nichole Watt

The equation for C is ${x}^{2}+{y}^{2}={r}^{2}$ and P is a point (a, 0)
on the x-axis with a $\ne ±r.$
The point on the circle C is(x, y).
Obtain the distance between (x, y) and (a, 0) as follows:
$d=\sqrt{\left(x-a{\right)}^{2}+\left(y-0{\right)}^{2}}$
$=\sqrt{\left(x-a{\right)}^{2}+{y}^{2}}$
Now, obtain the partial derivatives as follows,
${d}_{x}=0$
$\frac{2\left(x-a\right)}{2\sqrt{\left(x-a{\right)}^{2}+{y}^{2}}}=0$
$x=a$
${d}_{y}=0$
$\frac{2y}{2\sqrt{\left(x-a{\right)}^{2}+{y}^{2}}}=0$
$y=0$
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),
$d=\sqrt{\left(-r-a{\right)}^{2}+\left(0{\right)}^{2}}$
$=\sqrt{\left(r+a{\right)}^{2}}$
$=r+a$
and at (r, 0),
$d=\sqrt{\left(-r-a{\right)}^{2}+\left(0{\right)}^{2}}$
$=\sqrt{\left(r-a{\right)}^{2}}$
$=r-a$
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
and the minimum distance from the point P (a, 0) to the point
Hence, the maximum and minimum distances from P toa point X on C occur when the line XP goes through the center of C.