Question

Point P is the exterior of the circle with center O and radius is 15 units. A tangent is drawn from the point P that touches the circle at T. if PT is 8 units, then OP is:
A) 7 units
B) 13 units
C) 17 units
D) 23 units

Answer 1

Hint: Read the question carefully, and consider all the given information as it leads to a solution.
Tangent to a circle is a line that intersects the circle at only one point.
Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Figure: tangent XY to the circle with center O.
Hence, $OP \bot PQ$
Theorem 1 will be helpful in solving the given type of question.

Complete step-by-step answer:
Step 1: Draw the labeled diagram neatly.

Step 2: Use theorem 1
Given that a tangent from point P is drawn, and touches the given circle at point T.
The point of contact is point T.
OT is the radius of the given circle.
OT = 15.
By theorem 1, $OT \bot PT$
Step 3: Find OP using Pythagoras theorem
$\angle OTP = {90^ \circ }$ ,
The triangle formed by line joining points P, O, and T; is a right-angled triangle $\vartriangle POT$ right angle at T.

Thus, Pythagoras theorem can be applied on triangle $\vartriangle POT$ .
Pythagoras theorem: square of the hypotenuse is equal to the sum of the square of base and square of perpendicular.
 In a right angled triangle, base and perpendicular are at the angle of $90^\circ $ to each other and hypotenuse is the longest side.

  Figure 2: Right angled triangle ABC
$\mathop {{\text{Hypotenuse}}}\nolimits^{\text{2}} {\text{ = }}\mathop {{\text{ Base}}}\nolimits^{{\text{2 }}} {\text{ + }}\mathop {{\text{ Perpendicular}}}\nolimits^{\text{2}} $
Hence, side PT is a base , side OT is a perpendicular, side OP is a hypotenuse right-angled triangle $\vartriangle POT$.
Thus, $O{P^2} = O{T^2} + P{T^2}$
$
   \Rightarrow O{P^2} = {15^2} + {8^2} \\
   \Rightarrow O{P^2} = 225 + 64 \\
   \Rightarrow O{P^2} = 289 \\
   \Rightarrow OP = \sqrt {289} \\
  \because OP = 17 \\
 $

Final answer: The required length of the OP is 17 units. Thus, the correct option is (C).

Note: There is only one tangent at a point of contact of the circle.
Pythagoras theorem can be applied to every right angle triangle.
There is one another theorem related to tangents of the circle which can be useful for future reference.
Theorem 2: The lengths of tangents drawn from an external point to a circle are equal.

Figure: tangents PQ and PR to the circle with center O
Hence, length PQ = length PR