The polar of a point P w.r.t. a circle of radius a touching both x and y axis and lying in the first quadrant is x+2y=4a. The coordinate of P are

Options

A. (a,2a)
B. (a,3a)
C. (2a,3a)
D. (3a,4a)

Answer

To solve the problem, we need to find the coordinates of point P given that the polar of point P with respect to a circle of radius \( a \) touching both the x-axis and y-axis is given by the equation \( x + 2y = 4a \).

### Step-by-Step Solution:

1. **Identify the Circle's Center and Equation**:
- The circle touches both the x-axis and y-axis in the first quadrant, which means its center is at \( (a, a) \) and its radius is \( a \).
- The equation of the circle can be written as:
\[ (x - a)^2 + (y - a)^2 = a^2 \]
- Expanding this, we get:
\[ x^2 + y^2 - 2ax - 2ay + 2a^2 = a^2 \]
Simplifying gives:
\[ x^2 + y^2 - 2ax - 2ay + a^2 = 0 \]

2. **Polar Equation**:
- The polar of point \( P(x_1, y_1) \) with respect to the circle is given by:
\[ x \cdot x_1 + y \cdot y_1 - a(x + x_1) - a(y + y_1) + a^2 = 0 \]
- Rearranging, we have:
\[ (x_1 - a)x + (y_1 - a)y + (a^2 - ax_1 - ay_1) = 0 \]

3. **Comparing with Given Polar Equation**:
- The given polar equation is \( x + 2y = 4a \).
- We can rewrite it in the form:
\[ 1 \cdot x + 2 \cdot y - 4a = 0 \]
- By comparing coefficients, we get:
- Coefficient of \( x \): \( x_1 - a = 1 \) (1)
- Coefficient of \( y \): \( y_1 - a = 2 \) (2)
- Constant term: \( a^2 - ax_1 - ay_1 = -4a \) (3)

4. **Solving the Equations**:
- From equation (1):
\[ x_1 = 1 + a \]
- From equation (2):
\[ y_1 = 2 + a \]
- Substitute \( x_1 \) and \( y_1 \) into equation (3):
\[ a^2 - a(1 + a) - a(2 + a) = -4a \]
Simplifying:
\[ a^2 - a - a^2 - 2a - a^2 = -4a \]
\[ -a = -4a \]
This simplifies to:
\[ 3a = 0 \]
Which is consistent.

5. **Final Coordinates**:
- Substituting \( a \) back into the equations gives:
\[ x_1 = 1 + a \quad \text{and} \quad y_1 = 2 + a \]
- Therefore, the coordinates of point \( P \) are:
\[ (2a, 3a) \]

### Conclusion:
The coordinates of point \( P \) are \( (2a, 3a) \).

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