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

The equation for C is $x^2+y^2=r^2$ and P is a point (a, 0)
on the x-axis with a $\ne \pm r.$
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 $X(x,y)onCisr+a$
and the minimum distance from the point P (a, 0) to the point $X(x,y)onCisr-a.$
Hence, the maximum and minimum distances from P toa point X on C occur when the line XP goes through the center of C.