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# Differential equation of the family of circles touching the line $y=2$ at (0,2) is:A. ${{x}^{2}}+{{(y-2)}^{2}}+\dfrac{dy}{dx}(y-2)=0$B. ${{x}^{2}}+(y-2)\left( 2-2x\dfrac{dx}{dy}-y \right)=0$C. ${{x}^{2}}+{{(y-2)}^{2}}+\left( \dfrac{dx}{dy}+y-2 \right)(y-2)=0$D. None of the above

Last updated date: 25th Mar 2023
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Hint: Find the center and the radius of the circle from the equation of the circle with the given quantities. Differentiate them, find the value of k and substitute it in the equation of the circle, where k is the center of the circle.

We know the equation of a circle is ${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}........(1)$
Here the center of the family of circles will lie on the y-axis, so it can be taken of the form (0, k) where k is a constant.
Given the line $y=2$ at point (0, 2) it touches the circle.
Hence the radius of the circle lies from the center (0, k) to the point where the line touches at (0, 2). So by using the distance formula, we can find the radius of the circle.
Distance formula $=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}},$
Where $({{x}_{1}},{{y}_{1}})=(0,k)$ and$({{x}_{2}},{{y}_{2}})=(0,2)$,
Radius of circle = Distance between these 2 points,
\begin{align} & =\sqrt{{{(0-0)}^{2}}+{{(2-k)}^{2}}} \\ & =\sqrt{{{(2-k)}^{2}}}=2-k \\ \end{align}
Hence the radius of the circle is $2-k$ and center $(0,k)$. Substitute these values in the equation of the circle, i.e. in equation (1).
${{(x-0)}^{2}}+{{(y-k)}^{2}}={{(2-k)}^{2}}$
We know \begin{align} & {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\ & \Rightarrow {{x}^{2}}+({{y}^{2}}-2ky+{{k}^{2}})=4-4k+{{k}^{2}} \\ \end{align}
Cancel out the like terms,
\begin{align} & {{x}^{2}}+{{y}^{2}}-2ky+4k-4=0 \\ & {{x}^{2}}+{{y}^{2}}-2ky=4-4k......(2) \\ \end{align}
Now let us differentiate both sides of equation (2).
\begin{align} & \dfrac{d}{dx}({{x}^{2}}+{{y}^{2}}-2ky)=\dfrac{d}{dx}(4-4k) \\ & 2x+2y.\dfrac{dy}{dx}-2k.\dfrac{dy}{dx}=0-0 \\ \end{align}
Divide the expression by 2, we get,
\begin{align} & x+y.\dfrac{dy}{dx}-k.\dfrac{dy}{dx}=0 \\ & \therefore k.\dfrac{dy}{dx}=x+y.\dfrac{dy}{dx} \\ & k=\dfrac{x+y.\dfrac{dy}{dx}}{\dfrac{dy}{dx}}=x.\dfrac{dx}{dy}+y \\ & \therefore k=x.\dfrac{dx}{dy}+y.....(3) \\ \end{align}
Now let us go back to equation (2).
\begin{align} & {{x}^{2}}+{{y}^{2}}-2ky=4-4k \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-2ky-4-4k=0 \\ & {{x}^{2}}+{{y}^{2}}-2(ky+2-2k)=0.......(4) \\ \end{align}
Now let us substitute the value of k in equation (4) from equation (3).
${{x}^{2}}+{{y}^{2}}-2\left[ y\left( x.\dfrac{dx}{dy}+y \right)-2\left( x.\dfrac{dx}{dy}+y \right)+2 \right]=0$
Open the brackets and simplify the expression.
\begin{align} & {{x}^{2}}+{{y}^{2}}-2\left[ xy.\dfrac{dx}{dy}+{{y}^{2}}-2x.\dfrac{dx}{dy}-2y+2 \right]=0 \\ & {{x}^{2}}+{{y}^{2}}-2xy.\dfrac{dx}{dy}-2{{y}^{2}}+4x.\dfrac{dx}{dy}+4y-4=0 \\ & \left( {{x}^{2}}+{{y}^{2}}-2{{y}^{2}}+4y-4 \right)-2\left( xy.\dfrac{dx}{dy}-2x.\dfrac{dx}{dy} \right)=0 \\ & \left( {{x}^{2}}-{{y}^{2}}+4y-4 \right)-2\left( xy.\dfrac{dx}{dy}-2x\dfrac{dx}{dy} \right)=0 \\ & {{x}^{2}}-\left( {{y}^{2}}-4y+4 \right)-2x\dfrac{dx}{dy}(y-2)=0 \\ & \because {{y}^{2}}-4y+4={{(y-2)}^{2}}\left[ \because {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \right] \\ & \Rightarrow {{x}^{2}}-{{(y-2)}^{2}}-2x\dfrac{dx}{dy}(y-2)=0 \\ & {{x}^{2}}-(y-2)\left[ \left( y-2 \right)+2x\dfrac{dx}{dy} \right]=0 \\ & {{x}^{2}}+(y-2)\left[ 2-2x.\dfrac{dx}{dy}-y \right]=0 \\ \end{align}
Hence we got the differential equation of the family of circles touching the line $y=2$ at (0, 2).
Option D is the correct answer.

Note: It is said that the circle touches the line y=2, we have not been given the center of the circle but that it lies on y axis, x=0. So the center becomes (0, k) where k is a constant. So if a circle touches the line y=2, then the radius of the circle stretches from (0, k) to the line y=2.