The order of differential equation of family of circles passing through intersection of `3x+4y-7=0` and `S=-x^(2)+y^(2)-2x+1=0` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the order of the differential equation of the family of circles passing through the intersection of the line \(3x + 4y - 7 = 0\) and the circle given by the equation \(S = -x^2 + y^2 - 2x + 1 = 0\), we can follow these steps:

### Step 1: Understand the given equations
We have two equations:
1. The line: \(3x + 4y - 7 = 0\)
2. The circle: \(-x^2 + y^2 - 2x + 1 = 0\)

### Step 2: Formulate the family of circles
The family of circles that pass through the intersection of the line and the circle can be expressed as:
\[ -x^2 + y^2 - 2x + 1 + \lambda(3x + 4y - 7) = 0 \]
where \(\lambda\) is an arbitrary constant.

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ -x^2 + y^2 - 2x + 1 + 3\lambda x + 4\lambda y - 7\lambda = 0 \]
This can be rewritten as:
\[ -x^2 + y^2 + (3\lambda - 2)x + (4\lambda)y + (1 - 7\lambda) = 0 \]

### Step 4: Identify the arbitrary constant
In the above equation, we can see that \(\lambda\) is the only arbitrary constant present. The presence of one arbitrary constant indicates that the order of the differential equation is 1.

### Step 5: Conclusion
Thus, the order of the differential equation of the family of circles passing through the intersection of the given line and circle is:
\[ \text{Order} = 1 \]

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