If the straight line `ax + by = 2 ; a, b!=0`, touches the circle `x^2 +y^2-2x = 3` and is normal to the circle `x^2 + y^2-4y = 6`, then the values of 'a' and 'b' are ?

Options

A. (1,3)
B. (3,1)
C. (1,2)
D. (2,1)

Answer

To solve the problem, we need to find the values of 'a' and 'b' given the conditions about the line and the circles. Let's break down the solution step by step.

### Step 1: Identify the circles and their properties
The first circle is given by the equation:
\[ x^2 + y^2 - 2x = 3 \]
Rearranging this, we can write it as:
\[ (x - 1)^2 + y^2 = 4 \]
This shows that the center of the first circle is at \( (1, 0) \) and the radius \( r_1 = 2 \).

The second circle is given by the equation:
\[ x^2 + y^2 - 4y = 6 \]
Rearranging this, we can write it as:
\[ x^2 + (y - 2)^2 = 10 \]
This shows that the center of the second circle is at \( (0, 2) \) and the radius \( r_2 = \sqrt{10} \).

### Step 2: Find the perpendicular distance from the center of the first circle to the line
The line is given by:
\[ ax + by = 2 \]
The perpendicular distance \( d \) from the center of the first circle \( (1, 0) \) to the line is given by the formula:
\[ d = \frac{|a(1) + b(0) - 2|}{\sqrt{a^2 + b^2}} = \frac{|a - 2|}{\sqrt{a^2 + b^2}} \]
Since the line is a tangent to the first circle, this distance must equal the radius of the circle, which is 2:
\[ \frac{|a - 2|}{\sqrt{a^2 + b^2}} = 2 \]

### Step 3: Set up the equation from the distance condition
Squaring both sides, we get:
\[ |a - 2|^2 = 4(a^2 + b^2) \]
This can be expressed as:
\[ (a - 2)^2 = 4(a^2 + b^2) \]
Expanding both sides:
\[ a^2 - 4a + 4 = 4a^2 + 4b^2 \]
Rearranging gives us:
\[ 3a^2 + 4b^2 + 4a - 4 = 0 \]
This is our first equation.

### Step 4: Find the condition for the line being a normal to the second circle
Since the line is normal to the second circle, it must pass through the center of that circle \( (0, 2) \). Substituting \( x = 0 \) and \( y = 2 \) into the line equation:
\[ a(0) + b(2) = 2 \]
This simplifies to:
\[ 2b = 2 \]
Thus, we find:
\[ b = 1 \]

### Step 5: Substitute \( b \) back into the first equation
Now substituting \( b = 1 \) into the first equation:
\[ 3a^2 + 4(1)^2 + 4a - 4 = 0 \]
This simplifies to:
\[ 3a^2 + 4 + 4a - 4 = 0 \]
\[ 3a^2 + 4a = 0 \]
Factoring out \( a \):
\[ a(3a + 4) = 0 \]

### Step 6: Solve for \( a \)
From this equation, we have two possibilities:
1. \( a = 0 \) (which we discard since \( a \neq 0 \))
2. \( 3a + 4 = 0 \) leading to \( a = -\frac{4}{3} \)

### Conclusion
Thus, the values of \( a \) and \( b \) are:
\[ a = -\frac{4}{3}, \quad b = 1 \]