IIT-JEE 2007 · Paper 2 · Mathematics Q49

MathematicsSingle correct+3 / -1Question 49 of 66

The differential equation $\dfrac{dy}{dx}=\dfrac{\sqrt{1-y^2}}{y}$ determines a family of circles with

variable radii and a fixed centre at

(0,1)

variable radii and a fixed centre at

(0,-1)

fixed radius 1 and variable centres along the

x

-axis

fixed radius 1 and variable centres along the

y

-axis

Answer and worked solution

Official answer: C

Separate. $\dfrac{y\,dy}{\sqrt{1-y^2}}=dx$ , so $-\sqrt{1-y^2}=x+c$ .

Square. $(x+c)^2+y^2=1$ : circles of fixed radius 1 with centres $(-c,0)$ on the x-axis.

(C)

Topics:Variable separable Equation of a circle

IIT-JEE 2007 · Paper 2, Mathematics question 49