The differential equation of the family of circles passing through the origin and having centre at the line y=x is :

Solution

Maths sir · 4 steps

One circle, one unknown a: write the circle, differentiate once, then throw a out.

1. Write the family. A centre on y = x is (a, a). The circle passes through the origin, so its radius squared is 2a², and the circle is x² + y² - 2ax - 2ay = 0. There is one arbitrary constant, so the differential equation has order 1.
2. Differentiate once. Differentiating with respect to x gives 2x + 2yy' - 2a - 2ay' = 0, that is x + yy' = a(1 + y'). From the circle itself, a = (x² + y²)/2(x + y).

   2(x+y)(x + yy') = (x² + y²)(1 + y')

3. Collect terms. Expanding gives 2x² + 2xy + (2xy + 2y²)y' = x² + y² + (x² + y²)y'. Moving everything across gives x² + 2xy - y² = (x² - 2xy - y²) y'.
4. So. With y' = dy/dx, this reads (x² - y² + 2xy) dx = (x² - y² - 2xy) dy, which is option (d).

The idea to keep. Count the arbitrary constants first, since that is how many times you differentiate. Then use the original equation to write each constant in terms of x and y, and eliminate it.

Answer from the NTA final answer key. Solution by PrepWiser.