JEE Advanced 2015 · Paper 1 · Mathematics Q55

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MathematicsMultiple correct+4 / -2Question 55 of 60

Consider the family of all circles whose centers lie on the straight line $y=x$ . If this family of circles is represented by the differential equation $Py''+Qy'+1=0$ , where $P$ , $Q$ are functions of $x$ , $y$ and $y'$ (here $y'=\dfrac{dy}{dx}$ , $y''=\dfrac{d^2y}{dx^2}$ ), then which of the following statements is (are) true?

P=y+x

P=y-x

P+Q=1-x+y+y'+(y')^2

P-Q=x+y-y'-(y')^2

Answer and worked solution

Official answer: B, C

Family. $(x-h)^2+(y-h)^2=r^2$ .

First derivative. $(x-h)+(y-h)y'=0$ , so $h=\dfrac{x+yy'}{1+y'}$ and $y-h=\dfrac{y-x}{1+y'}$ .

Second derivative. $1+y'^2+(y-h)y''=0$ . Substituting $y-h$ :
$(y-x)y''+(1+y'+y'^2)y'+1=0.$

Read off. $P=y-x$ and $Q=1+y'+y'^2$ , so $P+Q=1-x+y+y'+y'^2$ .

(B), (C)

Topics:Formation and applications Family of circles

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JEE Advanced 2015 · Paper 1, Mathematics question 55