The differential equation of circles passing through the points of intersection of unit circle with centre at the origin and the line bisecting the first quadrant, is

Options

A. `y_(1)(x^(2)+y^(2)-1)+(x+yy_(1))=0`
B. `(y_(1)-1)(x^(2)+y^(2)-1)+(x+yy_(1))2(x-y)=0`
C. `y_(1)(x^(2)+y^(2)-1)+yy_(2) =0`
D. None of these

Answer

To find the differential equation of circles passing through the points of intersection of the unit circle with center at the origin and the line bisecting the first quadrant, we can follow these steps:

### Step 1: Identify the equations
The equation of the unit circle centered at the origin is:
\[ x^2 + y^2 = 1 \]
The equation of the line bisecting the first quadrant is:
\[ y = x \]

### Step 2: General equation of the circle
The general equation of a circle that passes through the points of intersection of the unit circle and the line can be expressed as:
\[ x^2 + y^2 - 1 + \lambda(y - x) = 0 \]
where \(\lambda\) is a parameter.

### Step 3: Differentiate the equation
To eliminate \(\lambda\), we differentiate the equation with respect to \(x\):
\[ \frac{d}{dx}(x^2 + y^2 - 1 + \lambda(y - x)) = 0 \]
Using the product rule and chain rule, we get:
\[ 2x + 2y \frac{dy}{dx} + (y - x) \frac{d\lambda}{dx} + \lambda \left(\frac{dy}{dx} - 1\right) = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2x + 2y \frac{dy}{dx} + \lambda \frac{dy}{dx} - \lambda = 0 \]
This can be rearranged to find \(\lambda\):
\[ \lambda = 2x + 2y \frac{dy}{dx} - 2y \frac{dy}{dx} \]

### Step 5: Substitute \(\lambda\) back into the original equation
Substituting \(\lambda\) back into the original circle equation gives:
\[ x^2 + y^2 - 1 + (2x + 2y \frac{dy}{dx} - 2y \frac{dy}{dx})(y - x) = 0 \]

### Step 6: Simplifying the equation
This leads to a complex equation, but we can simplify it to find the required differential equation of the circle. After simplification, we arrive at:
\[ y' - 1 + x^2 + y^2 - 1 - (2x + 2y)y' + (y - x) = 0 \]

### Final Step: Formulate the differential equation
The final differential equation of the circles passing through the points of intersection is:
\[ y' - 1 + x^2 + y^2 - 1 - (2x + 2y)y' + (y - x) = 0 \]