Question

Given that the slope of a tangent to the curve $y=y\left( x \right)$ y=y(x) at any point $\left( x,y \right)$ is $\dfrac{2y}{{{x}^{2}}}$ . If the curve passes through the centre of the circle ${{x}^{2}}+{{y}^{2}}-2x-2y=0$ then its equation is:
A. $x{{\log }_{e}}\left| y \right|=2\left( x-1 \right)$
B. $x{{\log }_{e}}\left| y \right|=x-1$
C. ${{x}^{2}}{{\log }_{e}}\left| y \right|=-2\left( x-1 \right)$
D. $x{{\log }_{e}}\left| y \right|=-2\left( x-1 \right)$

Answer 1

Hint: We will first start with finding the centre of the circle from the given equation, for this we will convert this given equation into the standard equation of the circle and then obtain the centre of the circle. After that we will write the slope of tangent in $\dfrac{dy}{dx}$ form and then integrate it in order to find the equation of the curve, since it passes through the centre of the circle we will put the values into the curve and find our final equation.

Complete step by step answer:
 First let’s find out the centre of the given circle:
We have: ${{x}^{2}}+{{y}^{2}}-2x-2y=0$ , Let’s arrange the terms with same variable together and rewrite the equation as: ${{x}^{2}}-2x+{{y}^{2}}-2y=0\Rightarrow \left( {{x}^{2}}-2x \right)+\left( {{y}^{2}}-2y \right)=0$
We will add and subtract one inside both the brackets as following:
$\left( {{x}^{2}}-2x+1-1 \right)+\left( {{y}^{2}}-2y+1-1 \right)=0\text{ }...........\text{ Equation 1}\text{.}$
Now we know that : ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Therefore, ${{x}^{2}}-2x+1$ in first bracket will be written as ${{\left( x-1 \right)}^{2}}$; Similarly for ${{y}^{2}}-2y+1$ in the second bracket will be written as ${{\left( y-1 \right)}^{2}}$ . Now we will put these value in equation 1:
$\begin{align}
  & \left( {{x}^{2}}-2x+1-1 \right)+\left( {{y}^{2}}-2y+1-1 \right)=0 \\
 & \Rightarrow \left( {{\left( x-1 \right)}^{2}}-1 \right)+\left( {{\left( y-1 \right)}^{2}}-1 \right)=0 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}-1+{{\left( y-1 \right)}^{2}}-1=0 \\
\end{align}$

Move both − 1 to the right side of the equation :
 \[\begin{align}
  & \Rightarrow {{\left( x-1 \right)}^{2}}-1+{{\left( y-1 \right)}^{2}}-1=0 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=1+1 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=2 \\
\end{align}\]
We know that this is the form of a circle : \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\] , We will use this form to determine the centre and radius of the circle.
We have \[{{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=2\], match the values in this circle to those of the standard form. The variable r represents the radius of the circle, h represents the x-offset from the origin, and k represents the y-offset from origin. The centre of the circle is found at (h,k)
Therefore, centre is $\left( 1,1 \right)$
 We are given the slope: $\dfrac{2y}{{{x}^{2}}}$ and we know that the slope are given by $\dfrac{dy}{dx}$ .
Therefore,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{2y}{{{x}^{2}}} \\
 & \Rightarrow \dfrac{dy}{2y}=\dfrac{dx}{{{x}^{2}}} \\
 & \Rightarrow \int{\dfrac{dy}{2y}}=\int{\dfrac{dx}{{{x}^{2}}}} \\
\end{align}$
To solve the given integration we will use the following properties:
\[\begin{align}
  & \int{\dfrac{1}{f\left( x \right)}dx=\ln f(x)} \\
 & \int{f{{(x)}^{n}}dx=\dfrac{f{{(x)}^{n+1}}}{n+1}} \\
\end{align}\]
Now we will solve the integral: $\int{\dfrac{dy}{2y}}=\int{\dfrac{dx}{{{x}^{2}}}}$
\[\begin{align}
  & \Rightarrow \int{\dfrac{dy}{2y}}=\int{\dfrac{dx}{{{x}^{2}}}} \\
 & \Rightarrow \dfrac{1}{2}\int{\dfrac{dy}{y}}=\int{\dfrac{dx}{{{x}^{2}}}} \\
 & \Rightarrow \dfrac{1}{2}{{\log }_{e}}y=\dfrac{{{x}^{(-2+1)}}}{\left( -2+1 \right)}+C \\
 & \Rightarrow \dfrac{1}{2}{{\log }_{e}}y=\dfrac{{{x}^{-1}}}{\left( -1 \right)}+C \\
 & \Rightarrow \dfrac{1}{2}{{\log }_{e}}y=\dfrac{-1}{x}+C \\
\end{align}\]
Since this curve passes through the centre: $(1,1)$ , it will satisfy the above equation:
Putting x=1 and y=1 in \[\dfrac{1}{2}{{\log }_{e}}y=\dfrac{-1}{x}+C\] we will have:
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}{{\log }_{e}}\left( 1 \right)=\dfrac{-1}{1}+C \\
 & \Rightarrow 0=-1+C \\
 & \Rightarrow C=1 \\
\end{align}\]
Putting C=1 in the obtained equation: \[\dfrac{1}{2}{{\log }_{e}}y=\dfrac{-1}{x}+C\], We will get the equation of curve as following:\[\]
\[\begin{align}
  & \dfrac{1}{2}{{\log }_{e}}y=\dfrac{-1}{x}+C \\
 & \Rightarrow \dfrac{1}{2}{{\log }_{e}}y=\dfrac{-1}{x}+1 \\
 & \Rightarrow \dfrac{1}{2}{{\log }_{e}}y=\dfrac{-1+x}{x}\Rightarrow x{{\log }_{e}}y=2\left( -1+x \right) \\
 & \Rightarrow x{{\log }_{e}}y=2\left( x-1 \right) \\
\end{align}\]
Now we will take the absolute value of y in order to cover the whole domain of the integral, therefore the equation of curve is as following:
\[x{{\log }_{e}}\left| y \right|=2\left( x-1 \right)\]
Hence, Option A is correct.

Note:
Remember to use the absolute value of log functions because we want to cover the whole domain. ${{\log }_{e}}y$ can also be written as $\ln y$, students should not get confused between these two, they mean the same. Chances of mistakes are higher while finding constants in the integral, so students must be conscious while putting the values of centre into the equation of the curve.