Two congruent circles with centers at (2, 3) and (5, 6), which intersects at right angles, have radius equal to
  A. $2\sqrt 3 $
  B. $3$
  C. $4$
  D. None

Hint: In order to solve the problem we will use distance formula to find the unknown points as it is mentioned in the question that centers and the intersecting point forms a right angled triangle. Further we will use Pythagoras formula here.

Complete step-by-step answer:

The figure according to the question is given above:
Let us assume $C_1$ is the center of the first circle which has coordinates (5, 6) and $C_2$ is the center of the second circle which has the coordinates (1, 2).
Now, in order to find the value of r. In the triangle $P{C_1}{C_2}$ , we must have the value of ${C_1}{C_2}$ .
As we know the distance formula.
If two points \[A\left( {{x_1},{y_1}} \right)\& B\left( {{x_2},{y_2}} \right)\] forms a line then distance of that line is given as:
\[D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
By applying the above formula the distance between points (2, 3) and (5, 6) is
\[
  {C_1}{C_2} = \sqrt {{{\left( {5 - 2} \right)}^2} + {{\left( {6 - 3} \right)}^2}} \\
   \Rightarrow {C_1}{C_2} = \sqrt {{{\left( 3 \right)}^2} + {{\left( 3 \right)}^2}} = \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2 \\
 \]
Since we know that the circles are congruent so the radius of both circles is the same.
Let r be the radius of both of the circles.
As provided in the problem statement that the congruent circles intersect at the right angle.
So triangle \[P{C_1}{C_2}\] is a right angled triangle with sides
\[P{C_1} = P{C_2} = r\] {radius are same}
Now, considering right angle triangle \[P{C_1}{C_2}\]
As we know according to Pythagoras theorem
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{sid}}{{\text{e}}_1}} \right)^2} + {\left( {{\text{sid}}{{\text{e}}_2}} \right)^2}$
So for triangle \[P{C_1}{C_2}\] we have
$
  {\left( {{{\text{C}}_1}{{\text{C}}_2}} \right)^2} = {\left( {{\text{P}}{{\text{C}}_1}} \right)^2} + {\left( {{\text{P}}{{\text{C}}_2}} \right)^2} \\
   \Rightarrow {\left( {3\sqrt 2 } \right)^2} = {\left( r \right)^2} + {\left( r \right)^2} \\
   \Rightarrow 2{\left( r \right)^2} = {\left( {3\sqrt 2 } \right)^2} = 9 \times 2 = 18 \\
   \Rightarrow {\left( r \right)^2} = \dfrac{{18}}{2} = 9 \\
   \Rightarrow r = \sqrt 9 = 3 \\
 $
Hence, the radius of the circle is equal to 3.
So, option B is the correct option.

Note: In order to solve such types of problems related to circles and other geometrical figures the formulas and concept of coordinate geometry must be remembered by students. Pythagoras theorem is one of the most important theorems and is used almost everywhere. Students must recognize the hypotenuse and the other sides first before applying this theorem.