JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q72

MathematicsSingle correct+4 / -1Question 72 of 90

The differential equation of the family of circles passing through the points (0, 2) and (0, −2) is :

2xy\dfrac{dy}{dx}+\left(x^2-y^2+4\right)=0

2xy\dfrac{dy}{dx}+\left(x^2+y^2-4\right)=0

2xy\dfrac{dy}{dx}+\left(y^2-x^2+4\right)=0

2xy\dfrac{dy}{dx}-\left(x^2-y^2+4\right)=0

Answer and worked solution

Official answer: A

Family $x^2+y^2+2gx-4=0$ . Differentiating, $g=-(x+yy')$ .

Eliminate. $2xyy'+x^2-y^2+4=0$ . (A)

Topics:Formation and applications Family of circles

JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 72