Smallest distance from a point P to any point on the circle C is 5 and the largest distance from the the point P to the circle is C is 11 . If point P is situated outside the circle C , then what is the distance between centre of the circle and the point P ?
MY View about problem ---------
11 is tangent to the circle C from point P
lets assume k is the point where tangent touch the circle C,
p is point outside circle X radius K touch point to tangent O centre of the circle so that ko is perpendicular to Kp by right angle triangle law po2= ko2+kp2 po=5+X Ko=X Kp=11 But answer is not matching with actual answer
Since the point $P$ is located outside of the circle, then we should draw a line, passing from $P$, and the center of the circle, $C$, whicih we call that line $l$. It is intuitive that $l$ intersects the circle in two points, $M$ and $N$ respectively. Then there are two segments: $\overline {PM}$ and $\overline {PN}$. According to the problem we have:
$|\overline {PM}|=5$ and $|\overline {PN}|=11$
So the diameter of the circle is equal to the difference of these values ( $\overline{MN}$ is one of the diameters of the circle), and hence:
$2R=11-5=6 \rightarrow R=3$
So the distance, between the center of the circle and $P$ is equal to:
$|\overline {CP}|=|\overline {PM}|+R=5+3=8$, in which $C$ is the center of the circle.
You can see it in the following picture:
Note that the tangent to the circle is not the longest distance from the point $P$, to the points of the circle, since if we draw the tangent and it intersects the circle in $S$, then we have the triangle $PSN$ which is $90$ degrees at the angle $S$, and we know the longest segment in a rectangular triangle is its hypotenuse, not the legs