Vertices of a variable triangle are `(3,4); (5costheta, 5sintheta)` and `(5sintheta,-5costheta)` where `theta` is a parameter then the locus of its orthocentre is a. `(x+y-1)^2+(x-y-7)^2=100` b. `(x+y-7)^2+(x-y-1)^2=100` c. `(x+y-7)^2+(x+y-1)^2=100` d. `(x+y-7)^2+(x-y+1)^2=100`

Options

A. `(x+y-1)^2+(x-y-7)^2=100`
B. `(x+y-7)^2+(x-y-1)^2=100`
C. `(x+y-7)^2+(x+y-1)^2=100`
D. `(x+y-7)^2+(x-y+1)^2=100`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Coordinates of the Vertices:**
- Vertex \( A \) is \( (3, 4) \)
- Vertex \( B \) is \( (5\cos\theta, 5\sin\theta) \)
- Vertex \( C \) is \( (5\sin\theta, -5\cos\theta) \)

2. **Calculate the Centroid \( G \) of the Triangle:**
The centroid \( G \) is given by the formula:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates:
\[ G\left( \frac{3 + 5\cos\theta + 5\sin\theta}{3}, \frac{4 + 5\sin\theta - 5\cos\theta}{3} \right) \]

3. **Express the Orthocenter \( H \):**
The orthocenter \( H \) of the triangle can be expressed in terms of the centroid \( G \) and the circumcenter \( O \). The circumcenter \( O \) is at the origin \( (0, 0) \). The centroid divides the line segment joining the orthocenter and circumcenter in the ratio \( 2:1 \).
Let the coordinates of the orthocenter be \( H(h, k) \). Then:
\[ G\left( \frac{h + 0}{3}, \frac{k + 0}{3} \right) = \left( \frac{h}{3}, \frac{k}{3} \right) \]

4. **Set Up the Equations:**
From the coordinates of the centroid:
- For the x-coordinate:
\[ \frac{h}{3} = \frac{3 + 5\cos\theta + 5\sin\theta}{3} \implies h = 3 + 5\cos\theta + 5\sin\theta \]
- For the y-coordinate:
\[ \frac{k}{3} = \frac{4 + 5\sin\theta - 5\cos\theta}{3} \implies k = 4 + 5\sin\theta - 5\cos\theta \]

5. **Express \( \sin\theta \) and \( \cos\theta \):**
Rearranging the equations:
- From \( h \):
\[ h - 3 = 5(\cos\theta + \sin\theta) \implies \cos\theta + \sin\theta = \frac{h - 3}{5} \]
- From \( k \):
\[ k - 4 = 5(\sin\theta - \cos\theta) \implies \sin\theta - \cos\theta = \frac{k - 4}{5} \]

6. **Add and Subtract the Equations:**
- Adding:
\[ \left( \frac{h - 3}{5} + \frac{k - 4}{5} \right) = 2\sin\theta \implies h + k - 7 = 10\sin\theta \]
- Subtracting:
\[ \left( \frac{h - 3}{5} - \frac{k - 4}{5} \right) = 2\cos\theta \implies h - k + 1 = 10\cos\theta \]

7. **Square and Add the Results:**
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[ \left( \frac{h + k - 7}{10} \right)^2 + \left( \frac{h - k + 1}{10} \right)^2 = 1 \]
Multiplying through by \( 100 \):
\[ (h + k - 7)^2 + (h - k + 1)^2 = 100 \]

8. **Substituting \( h \) and \( k \) with \( x \) and \( y \):**
Let \( h = x \) and \( k = y \):
\[ (x + y - 7)^2 + (x - y + 1)^2 = 100 \]

### Final Answer:
Thus, the locus of the orthocenter is given by:
\[ (x + y - 7)^2 + (x - y + 1)^2 = 100 \]