A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation `sqrt3 x+ y -6 = 0` and the point D is `((3sqrt3)/2, 3/2)`. Further, it is given that the origin and the centre of C are on the same side of the line PQ. (1)The equation of circle C is (2)Points E and F are given by (3)Equation of the sides QR, RP are

A. `y=(2)/(sqrt3)+x+1,y=-(2)/(sqrt3)x-1`
B. `y=(1)/(sqrt3)x,y=0`
C. `y=(sqrt3)/(2)x+1,y=-(sqrt3)/(2)x-1`
D. `y=sqrt3x,y=0`

Options

A. `y=(2)/(sqrt3)+x+1,y=-(2)/(sqrt3)x-1`
B. `y=(1)/(sqrt3)x,y=0`
C. `y=(sqrt3)/(2)x+1,y=-(sqrt3)/(2)x-1`
D. `y=sqrt3x,y=0`

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the necessary equations and coordinates.

### Step 1: Find the Equation of Circle C
Given that the radius of circle C is 1 and the center C is at coordinates \((\sqrt{3}, 1)\), the equation of the circle can be expressed using the standard form:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center and \(r\) is the radius. Substituting the values:

\[ (x - \sqrt{3})^2 + (y - 1)^2 = 1^2 \]

Thus, the equation of circle C is:

\[ (x - \sqrt{3})^2 + (y - 1)^2 = 1 \]

### Step 2: Find Points E and F
The points of contact D, E, and F are where the circle touches the sides of the triangle.

1. **Point D** is given as \(\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)\).
2. **Point F** lies on the side RP. Since the radius is 1 and the y-coordinate of point D is \(\frac{3}{2}\), the y-coordinate of F will be \(0\) (as it is on the x-axis). Thus, the coordinates of F can be determined as follows:

The x-coordinate of F will be the same as D, hence:
\[ F = \left(\frac{3\sqrt{3}}{2}, 0\right) \]

3. **Point E** lies on the side QR. The distance from C to E is also 1. Since the angle at C is 30 degrees, we can find E's coordinates using trigonometric relations:

The x-coordinate of E will be:
\[ E_x = \sqrt{3} + 1 \cdot \cos(30^\circ) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]
The y-coordinate of E will be:
\[ E_y = 1 + 1 \cdot \sin(30^\circ) = 1 + \frac{1}{2} = \frac{3}{2} \]

Thus, the coordinates of E are:
\[ E = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) \]

### Step 3: Find the Equations of Sides QR and RP
1. **Equation of Side QR**:
Using point E \(\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)\) and the slope of the line QR (which is \( \sqrt{3} \)), we can write the equation of the line in point-slope form:

\[ y - \frac{3}{2} = \sqrt{3}\left(x - \frac{3\sqrt{3}}{2}\right) \]

Simplifying this gives:
\[ y = \sqrt{3}x - \frac{3}{2} + \frac{3\sqrt{3}}{2} \]

Thus, the equation of QR is:
\[ y = \sqrt{3}x \]

2. **Equation of Side RP**:
The line RP is horizontal (parallel to the x-axis), so its equation is simply:
\[ y = 0 \]

### Final Answers
1. The equation of circle C is:
\[ (x - \sqrt{3})^2 + (y - 1)^2 = 1 \]
2. The coordinates of points E and F are:
\[ E = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right), \quad F = \left(\frac{3\sqrt{3}}{2}, 0\right) \]
3. The equations of the sides QR and RP are:
\[ y = \sqrt{3}x, \quad y = 0 \]