Maintaining Equal Distances

Relevant wiki: Cyclic Quadrilaterals

Clearly, that quadrilateral is a cyclic quadrilateral because $OAP$ and $OBP$ are both right angles since those are the angles between the radius and tangent lines. Hence their sum is $180^{\circ}$ , and from the angle sum property of a quadrilateral, the sum of the other two opposite sides is $180^{\circ}$ too.

Now, this cyclic quadrilateral can be circumscribed by a circle. Since these four points of the cyclic quadrilateral fall on the circle, the required equidistant point is the clearly the centre of the circle.

But just notice that the triangles $OBP$ and $OAP$ are right angled and since this right triangle also falls inside the circle with all its points on it, $OP$ is its diameter since the diameter always subtends a right angle anywhere on the circle. Therefore, the centre $M$ is the midpoint of the diametric line $OP$ , and can be found out using the midpoint formula:

$(\frac{3+9}{2}, \frac{5+11}{2}) = \boxed{(6,8)}$ .

Hence, distance from the origin $= \sqrt{6^2 + 8^2} = \boxed{10}$

PS: Sorry for the bad quality of the picture, went through a hard time circumscribing the quadrilateral.

Since the point is equidistant from all the points (VERICES) , THIS DISTANCE = R , THE RADIUS OF THE CIRCLE , i.e it will the mid-point of OP. So it's coordinates are (6,8). distance is √(6^2+8^2)=10.

The point which is equidistant from all the points will lie on perpendicular bisector of AB and OP. So it will the mid-point of OP. So it's coordinates are (6,8). Hence distance is √(6^2+8^2)=10.