The circle `x^(2)+y^(2)-2ax-2ay+a^(2)=0` touches axes of co ordinates at

Options

A. (a,a),(0,0)
B. (a,0),(0,0)
C. (a,0),(0,a)
D. (0,a),(1,a)

Answer

To determine the points where the circle given by the equation \( x^2 + y^2 - 2ax - 2ay + a^2 = 0 \) touches the coordinate axes, we can follow these steps:

### Step 1: Understand the Circle's Equation
The equation of the circle is given as:
\[ x^2 + y^2 - 2ax - 2ay + a^2 = 0 \]
This can be rearranged to identify the center and radius of the circle.

### Step 2: Rewrite the Circle's Equation
We can rewrite the equation in standard form by completing the square:
\[ (x^2 - 2ax + a^2) + (y^2 - 2ay + a^2) = 0 \]
This simplifies to:
\[ (x - a)^2 + (y - a)^2 = 0 \]
This indicates that the center of the circle is at the point \( (a, a) \) and the radius is \( 0 \), meaning it is a point circle.

### Step 3: Determine Touching Points on the X-axis
To find where the circle touches the x-axis, we set \( y = 0 \):
\[ (x - a)^2 + (0 - a)^2 = 0 \]
This simplifies to:
\[ (x - a)^2 + a^2 = 0 \]
Since the radius is \( 0 \), the only solution occurs when:
\[ x - a = 0 \implies x = a \]
Thus, the point of contact with the x-axis is:
\[ (a, 0) \]

### Step 4: Determine Touching Points on the Y-axis
Next, to find where the circle touches the y-axis, we set \( x = 0 \):
\[ (0 - a)^2 + (y - a)^2 = 0 \]
This simplifies to:
\[ a^2 + (y - a)^2 = 0 \]
Again, since the radius is \( 0 \), the only solution occurs when:
\[ y - a = 0 \implies y = a \]
Thus, the point of contact with the y-axis is:
\[ (0, a) \]

### Final Result
The circle touches the coordinate axes at the points:
\[ (a, 0) \text{ and } (0, a) \]