Question

In the Adjoining figure, if O is the centre of the circle of radius 5cm, OP=13 cm, where PQ and PR are tangents to the circle from point P. Length PR is :

• A. 10
• B. 12
• C. 15
• D. 17

Hint:

1. The radius to a point of tangency is always perpendicular to the tangent. So, \(\triangle \text{ORP}\) is a right-angled triangle at R. 2. In \(\triangle \text{ORP}\):
OR = radius = 5 cm (leg)
OP = distance from center = 13 cm (hypotenuse)
PR = length of tangent = ? (other leg) 3. Use Pythagoras theorem: \(OR^2 + PR^2 = OP^2\). \(5^2 + PR^2 = 13^2\) \(25 + PR^2 = 169\) \(PR^2 = 169 - 25 = 144\) \(PR = \sqrt{144} = 12\) cm. (This is a common Pythagorean triple: 5, 12, 13).

Answer 1

The Correct Option is B

Concept: (A) A tangent to a circle is perpendicular to the radius drawn to the point of tangency. (B) Tangents drawn from an external point to a circle are equal in length. (C) The Pythagorean theorem can be used in a right-angled triangle: \((\text{hypotenuse})^2 = (\text{perpendicular})^2 + (\text{base})^2\). Step 1: Identify the given information from the text and figure
O is the centre of the circle.
Radius of the circle (e.g., OR or OQ) = 5 cm.
Distance of point P from the centre O, OP = 13 cm.
PQ and PR are tangents to the circle from point P. R and Q are the points of tangency. We need to find the length of PR. Step 2: Apply the property that the radius is perpendicular to the tangent at the point of tangency Since PR is a tangent and OR is the radius to the point of tangency R, \(\triangle \text{ORP}\) is a right-angled triangle with the right angle at R (\(\angle \text{ORP} = 90^\circ\)). Similarly, \(\triangle \text{OQP}\) is a right-angled triangle with the right angle at Q (\(\angle \text{OQP} = 90^\circ\)). Step 3: Use the Pythagorean theorem in \(\triangle \text{ORP}\) In right-angled \(\triangle \text{ORP}\):
OR (radius) = 5 cm (one leg)
PR (tangent length) = ? (another leg)
OP (distance from centre to external point) = 13 cm (hypotenuse, as it's opposite the right angle) According to the Pythagorean theorem: \[ OR^2 + PR^2 = OP^2 \] Substitute the known values: \[ (5)^2 + PR^2 = (13)^2 \] \[ 25 + PR^2 = 169 \] Step 4: Solve for PR \[ PR^2 = 169 - 25 \] \[ PR^2 = 144 \] Take the square root of both sides: \[ PR = \sqrt{144} \] Since length must be positive: \[ PR = 12 \text{ cm} \] Step 5: Note about the length of PQ By the property that tangents from an external point to a circle are equal in length, PQ would also be equal to PR. So, PQ = 12 cm. The length of PR is 12 cm. This matches option (2).