In the given figure, O is the centre of the circle and PQ & PR are the tangents to the circle such that OQ = 8 cm and OP = 17 cm, then the length of the tangent PQ is :
Show Hint
This problem uses the well-known Pythagorean triplet \((8, 15, 17)\).
If you memorize basic triplets like \((3,4,5)\), \((5,12,13)\), and \((8,15,17)\), you can write down the answer of such geometry questions instantly without calculations.
Step 1: Understanding the Question:
This question is from Geometry, specifically the properties of circles and tangents.
We are given a circle with center O, and a tangent PQ drawn from an external point P.
We know the radius OQ and the distance OP from the center to the external point. We need to find the length of the tangent PQ.
Step 2: Key Formula or Approach:
A key theorem in circle geometry states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, \(\angle OQP = 90^{\circ}\).
This makes \(\triangle OQP\) a right-angled triangle with the hypotenuse being OP.
We can apply the Pythagoras Theorem:
\[ OP^2 = OQ^2 + PQ^2 \]
Step 3: Detailed Explanation:
In the right-angled triangle \(\triangle OQP\):
The radius \(OQ = 8\text{ cm}\).
The hypotenuse \(OP = 17\text{ cm}\).
Let the length of the tangent be \(PQ\).
By Pythagoras Theorem:
\[ OP^2 = OQ^2 + PQ^2 \]
Substitute the given values:
\[ 17^2 = 8^2 + PQ^2 \]
Calculate the squares:
\[ 289 = 64 + PQ^2 \]
Isolate \(PQ^2\):
\[ PQ^2 = 289 - 64 \]
\[ PQ^2 = 225 \]
Take the square root of both sides:
\[ PQ = \sqrt{225} \]
\[ PQ = 15\text{ cm} \]
Step 4: Final Answer:
The length of the tangent PQ is \(15\text{ cm}\).
