The tangents drawn from origin to the circle `x^2+y^2-2ax-2by+b^2` are perpendicular to each other, if a) `a-b =1` b) `a+b=1` c) `a^2-b^2 =0` d) `a^2+b^2=1`

Explanation

To determine the condition under which the tangents drawn from the origin to the circle \( x^2 + y^2 - 2ax - 2by + b^2 = 0 \) are perpendicular to each other, we can follow these steps:

### Step 1: Identify the center and radius of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 2ax - 2by + b^2 = 0 \]
We can rewrite this in the standard form:
\[ (x - a)^2 + (y - b)^2 = a^2 + b^2 - b^2 \]
From this, we can see that the center of the circle is \( (a, b) \) and the radius \( r \) is:
\[ r = \sqrt{a^2 + b^2 - b^2} = \sqrt{a^2} \]
Thus, the radius simplifies to:
\[ r = |a| \]

### Step 2: Use the condition for perpendicular tangents
The tangents from the origin to the circle are perpendicular if the product of their slopes is -1. The condition for the tangents from a point \( (x_0, y_0) \) to a circle \( (x - h)^2 + (y - k)^2 = r^2 \) is given by:
\[ \sqrt{(x_0 - h)^2 + (y_0 - k)^2} = r \]
For our case, substituting \( (x_0, y_0) = (0, 0) \):
\[ \sqrt{(0 - a)^2 + (0 - b)^2} = |a| \]
This leads to:
\[ \sqrt{a^2 + b^2} = |a| \]

### Step 3: Square both sides
Squaring both sides gives:
\[ a^2 + b^2 = a^2 \]
This simplifies to:
\[ b^2 = 0 \]
Thus, we find that:
\[ b = 0 \]

### Step 4: Substitute back to find conditions on \( a \)
Now substituting \( b = 0 \) back into the original conditions:
- For option (a) \( a - b = 1 \): \( a - 0 = 1 \) implies \( a = 1 \).
- For option (b) \( a + b = 1 \): \( a + 0 = 1 \) implies \( a = 1 \).
- For option (c) \( a^2 - b^2 = 0 \): \( a^2 - 0 = 0 \) implies \( a = 0 \).
- For option (d) \( a^2 + b^2 = 1 \): \( a^2 + 0 = 1 \) implies \( a = \pm 1 \).

### Conclusion
The only condition that holds true for the tangents to be perpendicular is when \( b = 0 \), leading to the conclusion that:
\[ \text{The correct answer is option (c): } a^2 - b^2 = 0. \]