If `H` is the othrocenter of an acute angled triangle ABC whose circumcircle is `x^2+y^2=16 ,` then circumdiameter of the triangle HBC is 1 (b) 2 (c) 4 (d) 8

Options

A. 1
B. 2
C. 4
D. 8

Explanation

To solve the problem step by step, we will analyze the given information and apply the properties of triangles and circles.

### Step-by-Step Solution:

1. **Identify the Circumradius**:
The circumcircle of triangle ABC is given by the equation \(x^2 + y^2 = 16\). This implies that the radius \(R\) of the circumcircle is:
\[ R = \sqrt{16} = 4. \]

**Hint**: Remember that the equation of a circle in the form \(x^2 + y^2 = r^2\) gives the radius \(r\).

2. **Understanding the Angles**:
In triangle ABC, the angles at vertices B and C are denoted as \(B\) and \(C\) respectively. The angle at the orthocenter \(H\) opposite to vertex \(A\) is \(B + C\).

**Hint**: The angles in a triangle sum up to \(180^\circ\) or \(\pi\) radians.

3. **Using the Sine Rule**:
According to the sine rule, the area \(A\) of triangle ABC can be expressed as:
\[ A = 2R \sin A. \]
Here, \(A\) is the angle at vertex A.

**Hint**: The sine rule relates the sides of a triangle to the sine of its angles.

4. **Finding the Circumradius of Triangle HBC**:
Let \(R_1\) be the circumradius of triangle \(HBC\). The area of triangle \(HBC\) can also be expressed as:
\[ A = 2R_1 \sin(B + C). \]

**Hint**: The circumradius of a triangle can be calculated using the area and sine of the angles.

5. **Relating Angles**:
Since \(B + C = \pi - A\), we can substitute this into the area formula:
\[ A = 2R_1 \sin(\pi - A) = 2R_1 \sin A. \]

**Hint**: The sine of an angle and its supplement are equal.

6. **Equating the Areas**:
From the two expressions for the area \(A\), we have:
\[ 2R \sin A = 2R_1 \sin A. \]
Dividing both sides by \(2 \sin A\) (assuming \(\sin A \neq 0\)), we find:
\[ R = R_1. \]

**Hint**: When dividing by a term, ensure it is not zero to avoid undefined expressions.

7. **Finding the Circumradius of Triangle HBC**:
Since we have established that \(R = R_1\) and \(R = 4\), it follows that:
\[ R_1 = 4. \]

**Hint**: The circumradius of triangle \(HBC\) is the same as that of triangle \(ABC\).

8. **Calculating the Circumdiameter**:
The circumdiameter \(D\) of triangle \(HBC\) is given by:
\[ D = 2R_1 = 2 \times 4 = 8. \]

**Hint**: The diameter is simply twice the radius.

### Final Answer:
The circumdiameter of triangle \(HBC\) is \(8\).