The equation(s) of the tangent at the point `(0,0)` to the circle, making intercepts of length `2a and 2b` units on the coordinate axes, is (are)

Options

A. `ax + by =0`
B. `ax -by =0`
C. `x =y`
D. none of these

Answer

To find the equation of the tangent at the point (0,0) to a circle that makes intercepts of length 2a and 2b on the coordinate axes, we can follow these steps:

### Step 1: Determine the intercepts
The tangent line makes intercepts of length 2a on the x-axis and 2b on the y-axis. This means that the x-intercept is at (2a, 0) and the y-intercept is at (0, 2b).

### Step 2: Write the equation of the tangent line
The general form of the equation of a line that intercepts the axes at (a, 0) and (0, b) is given by:
\[ \frac{x}{2a} + \frac{y}{2b} = 1 \]
To express this in a more standard form, we can multiply through by \(2ab\):
\[ bx + ay = 2ab \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ bx + ay - 2ab = 0 \]
This is the equation of the tangent line at the point (0,0).

### Step 4: Simplifying the equation
We can express this in a more recognizable form:
\[ ax + by = 2ab \]
This represents the equation of the tangent line in terms of the intercepts.

### Step 5: Conclusion
Thus, the equation of the tangent at the point (0,0) to the circle making intercepts of length 2a and 2b on the coordinate axes is:
\[ ax + by = 0 \]