Vertices of a variable triangle are `(3, 4), (5 cos theta, 5 sin theta)` and `(5 sin theta, -5 cos theta)`, where `theta in R`. Locus of its orthocentre is

Options

A. `x^(2)+y^(2)+6x+8y-25=0`
B. `x^(2)+y^(2)-6x+8y-25=0`
C. `x^(2)+y^(2)+6x-8y-25=0`
D. `x^(2)+y^(2)-6x-8y-25=0`

Answer

To find the locus of the orthocenter of the triangle with vertices at \( A(3, 4) \), \( B(5 \cos \theta, 5 \sin \theta) \), and \( C(5 \sin \theta, -5 \cos \theta) \), we can follow these steps:

### Step 1: Identify the coordinates of the vertices
The vertices of the triangle are given as:
- \( A(3, 4) \)
- \( B(5 \cos \theta, 5 \sin \theta) \)
- \( C(5 \sin \theta, -5 \cos \theta) \)

### Step 2: Find the circumcenter
We can observe that the circumradius (distance from the circumcenter to each vertex) is the same for all vertices. The distance from the origin \( O(0, 0) \) to each vertex can be calculated as follows:
- Distance \( OA = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = 5 \)
- Distance \( OB = \sqrt{(5 \cos \theta - 0)^2 + (5 \sin \theta - 0)^2} = \sqrt{25 (\cos^2 \theta + \sin^2 \theta)} = 5 \)
- Distance \( OC = \sqrt{(5 \sin \theta - 0)^2 + (-5 \cos \theta - 0)^2} = \sqrt{25 (\sin^2 \theta + \cos^2 \theta)} = 5 \)

Since \( OA = OB = OC = 5 \), the circumcenter \( O \) is at the origin \( (0, 0) \).

### Step 3: Find the centroid
The coordinates of the centroid \( G \) of triangle \( ABC \) can be calculated using the formula:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates of the vertices:
\[ G\left( \frac{3 + 5 \cos \theta + 5 \sin \theta}{3}, \frac{4 + 5 \sin \theta - 5 \cos \theta}{3} \right) \]

### Step 4: Use the relationship between the orthocenter, centroid, and circumcenter
The orthocenter \( H \) divides the line segment \( GO \) in the ratio \( 2:1 \). Therefore, we can express \( G \) in terms of \( H \) and \( O \):
\[ G = \frac{1H + 2O}{1 + 2} \]
Since \( O(0, 0) \), we have:
\[ G = \frac{1H + 2(0, 0)}{3} = \frac{H}{3} \]
Thus, we can write:
\[ G = \left( \frac{x}{3}, \frac{y}{3} \right) \]

### Step 5: Set up equations for \( H \)
From the coordinates of \( G \):
\[ \frac{x}{3} = \frac{3 + 5 \cos \theta + 5 \sin \theta}{3} \quad \text{and} \quad \frac{y}{3} = \frac{4 + 5 \sin \theta - 5 \cos \theta}{3} \]
This leads to:
\[ x = 3 + 5 \cos \theta + 5 \sin \theta \]
\[ y = 4 + 5 \sin \theta - 5 \cos \theta \]

### Step 6: Rearranging the equations
Rearranging gives:
\[ x - 3 = 5 \cos \theta + 5 \sin \theta \]
\[ y - 4 = 5 \sin \theta - 5 \cos \theta \]

### Step 7: Square both equations and add
Squaring both equations:
\[ (x - 3)^2 = 25(\cos \theta + \sin \theta)^2 \]
\[ (y - 4)^2 = 25(\sin \theta - \cos \theta)^2 \]

Adding these:
\[ (x - 3)^2 + (y - 4)^2 = 25 \left( (\cos \theta + \sin \theta)^2 + (\sin \theta - \cos \theta)^2 \right) \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ = 25 \left( 1 + 1 \right) = 50 \]

### Step 8: Final equation
Thus, we have:
\[ (x - 3)^2 + (y - 4)^2 = 50 \]

### Step 9: Rearranging to standard form
Expanding and rearranging gives:
\[ x^2 + y^2 - 6x - 8y - 25 = 0 \]

### Conclusion
The locus of the orthocenter is given by the equation:
\[ x^2 + y^2 - 6x - 8y - 25 = 0 \]