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
•
Total views: 305.4k
•
Views today: 4.82k
Answer
305.4k+ views
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.
“Complete step-by-step answer:”
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.
“Complete step-by-step answer:”
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.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
