The equations of the tangents drawn from the origin to the circle `x^2 + y^2 - 2px - 2qy + q^2 = 0` are perpendicular if (A) `p^2 = q^2` (B) `p^2 = q^2 =1` (C) `p = q/2` (D) `q= p/2`

Options

A. p=q
B. `p^(2) =q^(2)`
C. q= - p
D. `p^(2) + q^(2) =1 `

Answer

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

### Step 1: Identify the Circle's Center and Radius
The given equation of the circle is:
\[ x^2 + y^2 - 2px - 2qy + q^2 = 0 \]
We can rewrite it in standard form by completing the square.

The center \((h, k)\) of the circle can be identified as:
\[ h = p, \quad k = q \]
The radius \(r\) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
where \(g = -p\), \(f = -q\), and \(c = q^2\). Thus, we have:
\[ r = \sqrt{p^2 + q^2 - q^2} = \sqrt{p^2} = |p| \]

### Step 2: Equation of the Tangents from the Origin
The equation of the tangents from the origin \((0, 0)\) to the circle can be derived using the formula:
\[ y = mx \pm \sqrt{r^2(1 + m^2)} \]
where \(m\) is the slope of the tangent.

### Step 3: Condition for Perpendicular Tangents
For the tangents to be perpendicular, the product of their slopes must be \(-1\). If we denote the slopes of the tangents as \(m_1\) and \(m_2\), we have:
\[ m_1 \cdot m_2 = -1 \]

### Step 4: Finding the Slopes
Using the condition of tangents, we can derive that:
\[ m_1 = \frac{q}{p + \sqrt{p^2 + q^2}}, \quad m_2 = \frac{-q}{p - \sqrt{p^2 + q^2}} \]
Now, we need to find the condition under which:
\[ m_1 \cdot m_2 = -1 \]

### Step 5: Setting Up the Equation
Substituting \(m_1\) and \(m_2\) into the product:
\[ \left(\frac{q}{p + \sqrt{p^2 + q^2}}\right) \left(\frac{-q}{p - \sqrt{p^2 + q^2}}\right) = -1 \]
This simplifies to:
\[ \frac{-q^2}{(p + \sqrt{p^2 + q^2})(p - \sqrt{p^2 + q^2})} = -1 \]
Thus:
\[ q^2 = (p + \sqrt{p^2 + q^2})(p - \sqrt{p^2 + q^2}) \]

### Step 6: Solve the Equation
Expanding the right-hand side gives:
\[ q^2 = p^2 - (p^2 + q^2) = -q^2 \]
This leads to:
\[ 2q^2 = p^2 \]
or:
\[ p^2 = q^2 \]

### Conclusion
The condition for the tangents drawn from the origin to the circle to be perpendicular is:
\[ \boxed{p^2 = q^2} \]