If two circles and `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 of these

Answer

To solve the problem of determining the relationship between the coefficients of two circles that touch each other, we will follow these steps:

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

### Step 2: Rewrite the equations in standard form
We can rewrite the equations of the circles in standard form:
- For Circle 1:
\[ x^2 + y^2 + \frac{b}{a}x + \frac{c}{a}y = 0 \]
This gives us the center \( C_1 \left(-\frac{b}{2a}, -\frac{c}{2a}\right) \) and the radius \( r_1 = \sqrt{\left(-\frac{b}{2a}\right)^2 + \left(-\frac{c}{2a}\right)^2} = \frac{1}{2a}\sqrt{b^2 + c^2} \).

- For Circle 2:
\[ x^2 + y^2 + \frac{q}{p}x + \frac{r}{p}y = 0 \]
This gives us the center \( C_2 \left(-\frac{q}{2p}, -\frac{r}{2p}\right) \) and the radius \( r_2 = \sqrt{\left(-\frac{q}{2p}\right)^2 + \left(-\frac{r}{2p}\right)^2} = \frac{1}{2p}\sqrt{q^2 + r^2} \).

### Step 3: Use the condition for circles touching each other
For two circles to touch each other, the distance between their centers must equal the sum of their radii:
\[ C_1C_2 = r_1 + r_2 \]

### Step 4: Calculate the distance between the centers
The distance \( C_1C_2 \) is given by:
\[ C_1C_2 = \sqrt{\left(-\frac{b}{2a} + \frac{q}{2p}\right)^2 + \left(-\frac{c}{2a} + \frac{r}{2p}\right)^2} \]

### Step 5: Set up the equation
Setting the distance equal to the sum of the radii:
\[ \sqrt{\left(-\frac{b}{2a} + \frac{q}{2p}\right)^2 + \left(-\frac{c}{2a} + \frac{r}{2p}\right)^2} = \frac{1}{2a}\sqrt{b^2 + c^2} + \frac{1}{2p}\sqrt{q^2 + r^2} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides leads us to:
\[ \left(-\frac{b}{2a} + \frac{q}{2p}\right)^2 + \left(-\frac{c}{2a} + \frac{r}{2p}\right)^2 = \left(\frac{1}{2a}\sqrt{b^2 + c^2} + \frac{1}{2p}\sqrt{q^2 + r^2}\right)^2 \]

### Step 7: Simplify and derive the relationship
After simplification, we will find that:
\[ \frac{b}{q} = \frac{c}{r} \]
This implies that \( b \cdot r = c \cdot q \).

### Final Result
Thus, the condition for the two circles to touch each other is:
\[ b \cdot r = c \cdot q \]