Given that chords $PQ$ and $RS$ of a circle intersect at point $T$ at right angles, and the segments $PT = 2$ units, $TQ = 6$ units, and $TS = 3$ units, we are required to find the diameter of the circle.

Steps to Solve:

1. Use the Power of a Point Theorem: The power of a point theorem states that if two chords $AB$ and $CD$ intersect at point $T$ inside a circle, then: $PT \times TQ = RT \times TS$

   For our problem: $PT \times TQ = RT \times TS$

   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. Use the Formula for the Lengths of the Chords: The lengths of the chords can be related to the radius $R$ of the circle using the formula: $\text{Length of chord} = 2 \sqrt{R^2 - d^2}$ where $d$ is the perpendicular distance from the center to the chord.

   Since the chords intersect at right angles at $T$, and $PQ$ and $RS$ are perpendicular, we can apply Pythagoras’ theorem directly.

   $R^2 = \left(\frac{PQ}{2}\right)^2 + \left(\frac{RS}{2}\right)^2$ Substituting: $R^2 = \left(\frac{8}{2}\right)^2 + \left(\frac{7}{2}\right)^2 = 4^2 + 3.5^2$ $R^2 = 16 + 12.25 = 28.25$ $R = \sqrt{28.25} \approx 5.32 \text{ units}$

4. Find the Diameter: The diameter $D$ is given by: $D = 2R = 2 \times 5.32 \approx 10.64 \text{ units}$

Answer:

The diameter of the circle is approximately $\boxed{10.64}$ units.

---

Do you have any questions, or would you like a more detailed explanation?

Related Questions:

1. How would the solution change if the chords were not perpendicular?
2. What is the Power of a Point theorem, and how is it applied in different scenarios?
3. How do you find the radius if only one chord length is given?
4. How can the center of the circle be found using the chords?
5. What are other ways to find the diameter of a circle from intersecting chords?

Tip: Understanding the power of a point theorem is crucial as it often simplifies problems involving intersecting chords or tangents.