If the two circles `a(x^(2)+y^(2))+bx+cy=0` and `p(x^(2)+y^(2))+qx+ry=0` touch each other then

Options

A. `a//p=b//q`
B. `b//q=c//r`
C. `a//p=c//r`
D. none

Answer

To determine the condition under which the two circles \( a(x^2 + y^2) + bx + cy = 0 \) and \( p(x^2 + y^2) + qx + ry = 0 \) touch each other, we can follow these steps:

### Step 1: Identify the Circles
The given equations represent two circles:
1. Circle 1: \( a(x^2 + y^2) + bx + cy = 0 \)
2. Circle 2: \( p(x^2 + y^2) + qx + ry = 0 \)

### Step 2: Check if the Circles Pass Through the Origin
Both circles pass through the origin (0, 0) since substituting \( x = 0 \) and \( y = 0 \) satisfies both equations:
- For Circle 1: \( a(0^2 + 0^2) + b(0) + c(0) = 0 \)
- For Circle 2: \( p(0^2 + 0^2) + q(0) + r(0) = 0 \)

### Step 3: Find the Tangent Lines at the Origin
To find the tangent lines at the origin for both circles, we can use the formula for the tangent line at a point \((x_1, y_1)\) on a circle defined by \( T = S_1 \), where \( S_1 \) is the equation of the circle.

For Circle 1:
- The tangent line at the origin is derived from the equation:
\[ T_1 = a(0^2 + 0^2) + b(0) + c(0) = 0 \]
The tangent line simplifies to:
\[ bx + cy = 0 \]

For Circle 2:
- Similarly, the tangent line at the origin is:
\[ T_2 = p(0^2 + 0^2) + q(0) + r(0) = 0 \]
This simplifies to:
\[ qx + ry = 0 \]

### Step 4: Set the Tangent Lines Equal
Since the circles touch each other at the origin, the tangent lines at the origin must be the same. Therefore, we set the two equations equal:
\[ bx + cy = qx + ry \]

### Step 5: Compare Coefficients
For the two lines to be identical, the coefficients of \( x \) and \( y \) must be proportional:
\[ \frac{b}{q} = \frac{c}{r} \]

### Conclusion
The condition for the two circles to touch each other is:
\[ \frac{b}{q} = \frac{c}{r} \]