Given that chords $PQ$ and $RS$ of a circle intersect at point $T$ at right angles to each other, and the segments $PT = 2$ units, $TQ = 6$ units, and $TS = 3$ units respectively, we need to determine the diameter of the circle from the given options.

Steps to Solve:

1. Use the Power of a Point Theorem: The power of a point theorem states that if two chords intersect at a point, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord: $PT \times TQ = RT \times TS$ Substituting the given lengths: $2 \times 6 = RT \times 3$ $12 = 3 \times RT$ $RT = \frac{12}{3} = 4 \text{ units}$

2. Calculate the Lengths of the Chords:

   • $PQ = PT + TQ = 2 + 6 = 8$ units.
   • $RS = RT + TS = 4 + 3 = 7$ units.

3. Find the Length of the Diameter: Since the chords $PQ$ and $RS$ intersect at right angles, we can use the fact that the sum of the squares of the distances from $T$ to the ends of the chords equals the square of the diameter: $\text{Diameter}^2 = PT^2 + TQ^2 + RT^2 + TS^2$

   Substituting the values: $\text{Diameter}^2 = 2^2 + 6^2 + 4^2 + 3^2$ $\text{Diameter}^2 = 4 + 36 + 16 + 9$ $\text{Diameter}^2 = 65$ $\text{Diameter} = \sqrt{65} \text{ units}$

Answer:

The correct diameter of the circle is $\boxed{\sqrt{65}}$ units, corresponding to option (c).

---

Do you have any further questions, or would you like more details on any part of this solution?

Related Questions:

1. What is the significance of right-angle intersection in a circle's chords?
2. How does the power of a point theorem apply in different geometric configurations?
3. What is the relationship between the diameter and the chord lengths in a circle?
4. How can we find the radius if only the length of one chord is given?
5. What would be the approach if the chords did not intersect at right angles?

Tip: When solving geometry problems, always check if special theorems (like the Power of a Point) can simplify your calculations.