The equation of tangent drawn from origin to the circle `x^(2)+y^(2)-2ax-2by +b^(2)=0` are perpendicular if

Options

A. `a^(2)=b^(2)`
B. `a^(2)+b^(2)=1`
C. `2a=b`
D. `2b=a`

Answer

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

### Step 1: Identify the Circle's Center and Radius
The given equation of the circle can be rewritten in standard form. The general equation of a circle is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

From the equation \( x^2 + y^2 - 2ax - 2by + b^2 = 0 \), we can rearrange it to identify the center and radius:

\[ (x - a)^2 + (y - b)^2 = a^2 + b^2 - b^2 \]

Thus, the center of the circle \( C \) is \( (a, b) \) and the radius \( r \) is:

\[ r = \sqrt{a^2 + b^2 - b^2} = \sqrt{a^2} \]

So, the radius simplifies to \( r = |a| \).

### Step 2: Determine the Condition for Perpendicular Tangents
The tangents from the origin to the circle can be expressed as:

\[ y = mx \]

where \( m \) is the slope of the tangent. The tangents will be perpendicular if the product of their slopes is \( -1 \).

### Step 3: Find the Slopes of the Tangents
To find the slopes of the tangents from the origin to the circle, we can substitute \( y = mx \) into the circle's equation:

\[ x^2 + (mx)^2 - 2ax - 2b(mx) + b^2 = 0 \]

This simplifies to:

\[ (1 + m^2)x^2 - (2a + 2bm)x + b^2 = 0 \]

For tangents to exist, the discriminant of this quadratic must be zero:

\[ D = (2a + 2bm)^2 - 4(1 + m^2)b^2 = 0 \]

### Step 4: Set Up the Condition for Perpendicularity
From the discriminant condition, we can derive:

\[ (2a + 2bm)^2 = 4(1 + m^2)b^2 \]

For the tangents to be perpendicular, we need to find the condition under which the slopes \( m_1 \) and \( m_2 \) satisfy \( m_1 \cdot m_2 = -1 \).

### Step 5: Solve for the Condition
From the geometry of the problem, we can see that if the tangents are perpendicular, the distance from the center of the circle to the origin must equal the radius. This gives us the condition:

\[ \sqrt{a^2 + b^2} = |a| \]

Squaring both sides, we get:

\[ a^2 + b^2 = a^2 \]

This implies:

\[ b^2 = 0 \quad \Rightarrow \quad b = 0 \]

### Conclusion
Thus, the condition for the tangents drawn from the origin to the circle \( x^2 + y^2 - 2ax - 2by + b^2 = 0 \) to be perpendicular is:

\[ a^2 = b^2 \]