JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
The differential equation of the family of circles passing through the points (0,2) and (0,-2) is :
- (a) 2 x y (d y)/(d x)+(x²-y²+4)=0
- (b) 2 x y (d y)/(d x)+(x²+y²-4)=0
- (c) 2 x y (d y)/(d x)+(y²-x²+4)=0
- (d) 2 x y (d y)/(d x)-(x²-y²+4)=0
Answer and solution
Answer (a) 2 x y (d y)/(d x)+(x²-y²+4)=0
Solution
Maths sir · 4 steps
Both points sit on the y-axis, so every centre lies on the x-axis and only one constant is left to remove.
- Set up the family. The centre must lie on the perpendicular bisector of the two points, which is the x-axis. So the circles are x² + y² + 2gx + c = 0. Putting in (0, 2) gives 4 + c = 0, so c = -4.
x² + y² + 2gx - 4 = 0
- Differentiate once. There is one arbitrary constant g, so one differentiation is enough. This gives 2x + 2ydy/dx + 2g = 0, so 2g = -(2x + 2ydy/dx).
- Eliminate g. Substitute into the family: x² + y² - 2x² - 2xydy/dx - 4 = 0, which is -x² + y² - 2xydy/dx - 4 = 0. Multiply through by -1.
- So. The equation is 2xydy/dx + (x² - y² + 4) = 0, which is option (a).
The idea to keep. Count the arbitrary constants left after using the given conditions. That count is the order. Differentiate that many times, then eliminate every constant.
Answer from the NTA final answer key. Solution by PrepWiser.