JEE Main 202228 Jul, Evening shiftMathematicsEasyDifferential Equations163× since 2002The differential equation of the family of circles passing through the points (0,2)(0,2)(0,2) and (0,−2)(0,-2)(0,−2) is :AOption A. 2xydydx+(x2−y2+4)=02 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=02xydxdy+(x2−y2+4)=0BOption B. 2xydydx+(x2+y2−4)=02 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=02xydxdy+(x2+y2−4)=0COption C. 2xydydx+(y2−x2+4)=02 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=02xydxdy+(y2−x2+4)=0DOption D. 2xydydx−(x2−y2+4)=02 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=02xydxdy−(x2−y2+4)=0CheckSkip · show solution
