Question 3

Tangent: A line that touches a circle at exactly one point (the point of tangency) without crossing through it. The radius drawn to that point is always perpendicular to the tangent.

Let's use the property that the radius is perpendicular to the tangent at the point of contact.

Step 1 — Identify the given information

We have a circle. Its center is O. Its radius OP is 5 cm. PQ is a tangent to the circle. P is the point of tangency. A line from the center O meets the tangent at Q. The length OQ is 12 cm. We need to find the length of PQ.

Step 2 — Form a right-angled triangle

Tangent-Radius Perpendicularity Theorem: The radius drawn to the point of tangency is always perpendicular to the tangent at that point. This is a fundamental property of tangents.

The radius OP is perpendicular to the tangent PQ. This happens at the point of contact P. So, angle OPQ is $90^\circ$. Triangle OPQ is a right-angled triangle. OP and PQ are the legs of this triangle. OQ is the hypotenuse — it is the side opposite the 90° angle at P, and is always the longest side.

Step 3 — Apply Pythagoras Theorem

Pythagoras Theorem: In a right-angled triangle, $\text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2$.

We can use the Pythagoras theorem. The square of the hypotenuse equals the sum of the squares of the other two sides. In $\triangle OPQ$, $OQ^2 = OP^2 + PQ^2$. Let's substitute the known values.

$12^2 = 5^2 + PQ^2$

$144 = 25 + PQ^2$

$PQ^2 = 144 - 25$

$PQ^2 = 119$

$PQ = \sqrt{119}$

$\boxed{\text{PQ} = \sqrt{119} \text{ cm}}$

Answer

(D) $\sqrt{119}$ cm.