JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q78

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MathematicsSingle correct+4 / -1Question 78 of 90

If the differential equation representing the family of all circles touching $x$ -axis at the origin is $\left(x^2-y^2\right)\dfrac{dy}{dx}=g(x)\,y$ , then $g(x)$ equals :

\dfrac12x

2x^2

2x

\dfrac12x^2

Answer and worked solution

Official answer: C

The circles are $x^2+y^2=2ky$ . Differentiating, $x+yy'=ky'$ , so $k=\dfrac{x+yy'}{y'}$ . Substituting:
$(x^2+y^2)y'=2y(x+yy')\implies(x^2-y^2)y'=2xy.$
So $g(x)=2x$ .

Topics:Formation and applications Equation of a circle

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 78