Suppose the solution of the differential equation (d y)/(d x)=((2+α) x-β y+2)/(β x-2 α y-(β γ-4 α)) represents a circle passing through origin. Then the radius of this circle is :

Solution

Maths sir · 4 steps

Cross-multiply, spot the exact differentials, then force the integrated curve to be a circle.

1. Cross-multiply. Write it as [(2+α)x - β y + 2] dx = [β x - 2α y - (βγ - 4α)] dy and bring everything to one side. The terms -β y dx and -β x dy combine into -β d(xy), and every other term is a function of x alone or y alone.

   (2+α)/2x² + 2x - β xy + α y² + (βγ - 4α)y = C

2. Make it a circle. A circle has no xy term, so β = 0. The x² and y² coefficients must be equal, so (2+α)/2 = α, giving α = 2. It passes through the origin, so C = 0.
3. Read the circle. The curve becomes 2x² + 2x + 2y² - 8y = 0, that is, x² + y² + x - 4y = 0. The centre is (-1/2, 2) and r = √(1/4 + 4) = √(17/4).
4. So. The radius is √17/2, which is option (c).

The idea to keep. When the numerator and denominator are both linear, cross-multiply and look for d(xy). For a circle, remove the xy term and make the x² and y² coefficients equal.

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