JEE Advance - Mathematics (2015 - Paper 1 Offline - No. 11)
Consider the family of all circles whose centres lie on the straight line $$y=x,$$ If this family of circle is represented by the differential equation $$Py'' + Qy' + 1 = 0,$$ where $$P, Q$$ are functions of $$x,y$$ and $$y'$$ $$\left( {here\,\,\,y' = {{dy} \over {dx}},y'' = {{{d^2}y} \over {d{x^2}}}} \right)$$ then which of the following statements is (are) true?
$$P = y + x$$
$$\,P = y - x$$
$$\,P + Q = 1 - x + y + y' + {\left( {y'} \right)^2}$$
$$\,P - Q = 1 - x + y - y' - {\left( {y'} \right)^2}$$
Explanation
Let equation of circle whose centre lie on straight line $y=x$ be
$(x-k)^2+(y-k)^2=r^2~~~~...(i)$
Differentiating the above equation w.r.t. $x$, we get
$$ \begin{aligned} & 2(x-k)+2(y-k) y^{\prime}=0 \\\\ \Rightarrow & x+y y^{\prime}=k\left(1+y^{\prime}\right)~~~~~...(ii) \\\\ \Rightarrow & k=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned} $$
Differentiating the eq. (ii) w.r.t. $x$, we get
$$ \begin{aligned} & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=k y^{\prime \prime} \quad\left\{\because(u v)^{\prime}=u v^{\prime}+v u^{\prime}\right\} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right) y^{\prime \prime} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2+y^{\prime}+y y^{\prime} y^{\prime \prime}+\left(y^{\prime}\right)^3 \\\\ & =x y^{\prime \prime}+y y^{\prime} y^{\prime \prime} \\\\ & \Rightarrow y^{\prime \prime}(y-x)+\left(y^{\prime}\right)^2\left(1+y^{\prime}\right)+1+y^{\prime}=0 \\\\ & \Rightarrow y^{\prime \prime}(y-x)+y^{\prime}\left(y^{\prime}+\left(y^{\prime}\right)^2+1\right)+1=0 \end{aligned} $$
Comparing the above equation with $p y^{\prime \prime}+\mathrm{Q} y^{\prime}$ $+1=0$, we get
$$ \begin{aligned} & P=y-x \text { and } \mathrm{Q}=y^{\prime}+\left(y^{\prime}\right)^2+1 \\\\ \therefore & P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2 \end{aligned} $$
$(x-k)^2+(y-k)^2=r^2~~~~...(i)$
Differentiating the above equation w.r.t. $x$, we get
$$ \begin{aligned} & 2(x-k)+2(y-k) y^{\prime}=0 \\\\ \Rightarrow & x+y y^{\prime}=k\left(1+y^{\prime}\right)~~~~~...(ii) \\\\ \Rightarrow & k=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned} $$
Differentiating the eq. (ii) w.r.t. $x$, we get
$$ \begin{aligned} & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=k y^{\prime \prime} \quad\left\{\because(u v)^{\prime}=u v^{\prime}+v u^{\prime}\right\} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right) y^{\prime \prime} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2+y^{\prime}+y y^{\prime} y^{\prime \prime}+\left(y^{\prime}\right)^3 \\\\ & =x y^{\prime \prime}+y y^{\prime} y^{\prime \prime} \\\\ & \Rightarrow y^{\prime \prime}(y-x)+\left(y^{\prime}\right)^2\left(1+y^{\prime}\right)+1+y^{\prime}=0 \\\\ & \Rightarrow y^{\prime \prime}(y-x)+y^{\prime}\left(y^{\prime}+\left(y^{\prime}\right)^2+1\right)+1=0 \end{aligned} $$
Comparing the above equation with $p y^{\prime \prime}+\mathrm{Q} y^{\prime}$ $+1=0$, we get
$$ \begin{aligned} & P=y-x \text { and } \mathrm{Q}=y^{\prime}+\left(y^{\prime}\right)^2+1 \\\\ \therefore & P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2 \end{aligned} $$
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