Note: The option should be circumcentre instead of circumference
Let OP be the pole inside the triangular park ABC. Here ∠OAP = ∠OBP = ∠OCP = θ (Say) In ΔAOP, tanθ=
OP
OA
⇒OA=
OP
tanθ
...(1) In right ΔBOP, tanθ=
OP
OB
⇒OB=
OP
tanθ
....(2) In right ΔCOP, tanθ=
OP
OC
⇒OC=
OP
tanθ
...(3) From (1), (2) and (3), we have OA = OB = OC So, O is equidistant from the vertices of the ΔABC. Thus, O is the circumcentre of ΔABC. Therefore, the foot of the pole is at the circumcentre.