A circle C whose radius is 1 unit, thuches the x-axis at point A. The centre Q of C lies in first quadrant. The tangent from origin O to the circle touches it at T and a point P lies on it such that `Delta OAP` is a right angled triangle at A and its perimeter is 8 units.
Equation of tangent OT is

Options

A. `x - sqrt3 y =0`
B. `x - sqrt2 y =0`
C. `y - sqrt3 x =0`
D. none of these

Answer

To solve the problem step by step, we will follow the given information and derive the required equation of the tangent line from the origin to the circle.

### Step 1: Define the Circle
The circle \( C \) has a radius of 1 unit and touches the x-axis at point \( A \). Since the center \( Q \) of the circle lies in the first quadrant, we can denote the coordinates of point \( A \) as \( (h, 0) \) and the coordinates of the center \( Q \) as \( (h, 1) \).

### Step 2: Determine the Coordinates of Point \( P \)
Point \( P \) lies on the tangent line from the origin \( O(0, 0) \) to the circle at point \( T \). The triangle \( OAP \) is a right-angled triangle at \( A \), which means the angle \( OAP \) is \( 90^\circ \).

### Step 3: Write the Equation of the Tangent Line
The equation of the tangent line \( OT \) can be expressed using the slope formula. If we denote the coordinates of point \( P \) as \( (h, \alpha) \), the slope of line \( OP \) is given by:
\[ \text{slope} = \frac{\alpha - 0}{h - 0} = \frac{\alpha}{h} \]
Thus, the equation of the line can be written as:
\[ y = \frac{\alpha}{h}x \]
Rearranging gives:
\[ \alpha x - hy = 0 \tag{1} \]

### Step 4: Find the Perpendicular Distance from the Center to the Tangent
The perpendicular distance from the center \( Q(h, 1) \) to the tangent line \( \alpha x - hy = 0 \) must equal the radius of the circle, which is 1 unit. The formula for the distance \( d \) from point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our tangent line, \( A = \alpha \), \( B = -h \), and \( C = 0 \). Thus, the distance from \( Q(h, 1) \) is:
\[ d = \frac{|\alpha h - h \cdot 1|}{\sqrt{\alpha^2 + h^2}} = \frac{|\alpha h - h|}{\sqrt{\alpha^2 + h^2}} = \frac{h|\alpha - 1|}{\sqrt{\alpha^2 + h^2}} = 1 \]

### Step 5: Solve for \( \alpha \)
Squaring both sides gives:
\[ h^2(\alpha - 1)^2 = \alpha^2 + h^2 \]
Expanding and rearranging leads to:
\[ h^2(\alpha^2 - 2\alpha + 1) = \alpha^2 + h^2 \]
\[ h^2\alpha^2 - 2h^2\alpha + h^2 = \alpha^2 + h^2 \]
\[ (h^2 - 1)\alpha^2 - 2h^2\alpha = 0 \]
Factoring out \( \alpha \):
\[ \alpha((h^2 - 1)\alpha - 2h^2) = 0 \]
Thus, \( \alpha = 0 \) or \( (h^2 - 1)\alpha = 2h^2 \) leading to:
\[ \alpha = \frac{2h^2}{h^2 - 1} \]

### Step 6: Use the Perimeter Condition
The perimeter of triangle \( OAP \) is given as 8 units:
\[ OA + AP + OP = 8 \]
Substituting the lengths:
\[ h + \alpha + \sqrt{h^2 + \alpha^2} = 8 \]
Substituting \( \alpha \):
\[ h + \frac{2h^2}{h^2 - 1} + \sqrt{h^2 + \left(\frac{2h^2}{h^2 - 1}\right)^2} = 8 \]
This can be simplified further to find \( h \).

### Step 7: Solve for \( h \)
After solving the cubic equation derived from the perimeter condition, we find \( h = 2 \).

### Step 8: Calculate \( \alpha \)
Substituting \( h = 2 \) into the equation for \( \alpha \):
\[ \alpha = \frac{2(2^2)}{2^2 - 1} = \frac{8}{3} \]

### Step 9: Final Equation of the Tangent
Substituting \( h \) and \( \alpha \) back into the tangent line equation:
\[ y = \frac{8/3}{2}x \implies y = \frac{4}{3}x \]
Rearranging gives:
\[ 4x - 3y = 0 \]

### Final Answer
The equation of the tangent \( OT \) is:
\[ 4x - 3y = 0 \]