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 will follow these steps:

### Step 1: Find the Circumcenter
The circumcenter \( O \) of the triangle formed by the points \( A \), \( B \), and \( C \) is the point equidistant from all three vertices.

1. Calculate the distances from the origin \( O(0, 0) \) to each vertex:
- \( OA = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
- \( OB = \sqrt{(5 \cos \theta - 0)^2 + (5 \sin \theta - 0)^2} = \sqrt{25 (\cos^2 \theta + \sin^2 \theta)} = \sqrt{25} = 5 \)
- \( OC = \sqrt{(5 \sin \theta - 0)^2 + (-5 \cos \theta - 0)^2} = \sqrt{25 (\sin^2 \theta + \cos^2 \theta)} = \sqrt{25} = 5 \)

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

### Step 2: Find the Centroid
The centroid \( G \) of the triangle 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 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 3: Relationship Between Orthocenter and Centroid
The orthocenter \( H \) of a triangle is related to the centroid \( G \) and the circumcenter \( O \) by the formula:
\[ HG = \frac{2}{3} GO \]
Since \( O \) is at the origin \( (0, 0) \), we can express \( H \) in terms of \( G \):
\[ H = G + \frac{2}{3} (O - G) = G - \frac{2}{3} G + \frac{2}{3} O = \frac{1}{3} G + \frac{2}{3} O \]
This simplifies to:
\[ H = \frac{1}{3} G \]

### Step 4: Substitute for \( G \)
Substituting \( G \) into the equation for \( H \):
\[ H = \frac{1}{3} \left( \frac{3 + 5 \cos \theta + 5 \sin \theta}{3}, \frac{4 + 5 \sin \theta - 5 \cos \theta}{3} \right) \]
This gives:
\[ H = \left( \frac{3 + 5 \cos \theta + 5 \sin \theta}{9}, \frac{4 + 5 \sin \theta - 5 \cos \theta}{9} \right) \]

### Step 5: Find the Locus of \( H \)
To find the locus, we eliminate \( \theta \). Let:
\[ x = \frac{3 + 5 \cos \theta + 5 \sin \theta}{9}, \quad y = \frac{4 + 5 \sin \theta - 5 \cos \theta}{9} \]
Rearranging gives:
\[ 9x = 3 + 5 \cos \theta + 5 \sin \theta \quad \text{and} \quad 9y = 4 + 5 \sin \theta - 5 \cos \theta \]
From these, we can express \( \cos \theta \) and \( \sin \theta \) in terms of \( x \) and \( y \) and then substitute to find a relationship between \( x \) and \( y \).

### Step 6: Final Equation
After substituting and simplifying, we arrive at the equation of the locus:
\[ (x - 3)^2 + (y - 4)^2 = 25 \]
This represents a circle with center \( (3, 4) \) and radius \( 5 \).