Answer
Verified
493.8k+ views
Hint: The equation of the sphere passes through the intersection of circle and the plane. Substitute the equation of sphere and plane in\[\left( S+\lambda L \right)\]. Find the center of the sphere and compare it with the general equation of the sphere. By putting the center in the equation of the plane to get the greater circle, find \[\lambda \].
Complete step-by-step answer:
Given the equation of circle as \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5=0\]
The equation of the plane is given as \[5x-2y+4z+7=0\].
In this question, we need to find the equation of the sphere.
The equation of the sphere is passing through the point of intersection of the sphere and the plane. Thus the equation of sphere can be said as \[\left( S+\lambda L \right)\],
Where S = equation of circle =\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5=0\]
L = equation of plane =\[5x-2y+4z+7=0\] and \[\lambda \]is constant.
\[S+\lambda L=\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5 \right)+\lambda \left( 5x-2y+4z+7 \right)=0\]
\[={{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( 3-5\lambda \right)x+\left( 4-2\lambda \right)y-\left( 2-4\lambda \right)z-\left( 5+7\lambda \right)=0......(1)\]
We know the general equation of sphere as,
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2wy+2vz+d=0.....(2)\]
Now let us compare equation (1) and equation (2).
From that we get,
\[\begin{align}
& 2u=-\left( 3-5\lambda \right)=5\lambda -3 \\
& 2w=\left( 4-2\lambda \right) \\
& 2v=-\left( 2-4\lambda \right)=4\lambda -2 \\
\end{align}\]
\[u=\dfrac{5\lambda -3}{2},w=\dfrac{4-2\lambda }{2}=2-\lambda ,v=\dfrac{4\lambda -2}{2}=2\lambda -1\]
Thus, the center of sphere is\[\left( -u,-v,-z \right)\]
\[=\left[ -\left( \dfrac{5\lambda -3}{2} \right),-\left( 2-\lambda \right),-\left( 2\lambda -1 \right) \right]\]
\[=\left[ \dfrac{3-5\lambda }{2},\lambda -2,1-2\lambda \right]......(3)\]
The center of the sphere lies on the plane if it is a great circle.
Therefore, let us substitute these values in the equation of the plane.
\[\begin{align}
& 5x-2y+4z+7=0 \\
& 5\left( \dfrac{3-5\lambda }{2} \right)-2\left( \lambda -2 \right)+4\left( 1-2\lambda \right)+7=0 \\
\end{align}\]
Open the brackets and simplify the expression to get the value of\[\lambda \].
\[=\dfrac{15}{2}-\dfrac{25\lambda }{2}-2\lambda +4+4-8\lambda +7=0\]
\[=\left( \dfrac{15}{2}+4+4+7 \right)-\lambda \left( \dfrac{25}{2}+2+8 \right)=0\]
\[=\left( \dfrac{15+8+8+14}{2} \right)-\lambda \left( \dfrac{25+4+16}{2} \right)=0\]
\[=\dfrac{45}{2}-\lambda \left( \dfrac{45}{2} \right)=0\]
\[\begin{align}
& =\dfrac{45-45\lambda }{2}=0 \\
& \therefore 45-45\lambda =0 \\
& \therefore 45=45\lambda \\
& \therefore \lambda =1 \\
\end{align}\]
Hence we got the value of constant\[\lambda =1\].
Now put the value of\[\lambda =1\] in equation (1).
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( 3-5\times 1 \right)x+\left( 4-2\times 1 \right)y-\left( 2-4\times 1 \right)z-\left( 5+7\times 1 \right)=0 \\
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( -2 \right)x+2y-\left( -2 \right)z-12=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2x+2y+2z-12=0.....(4) \\
\end{align}\]
Thus, we got the equation of the sphere.
Now, put the value of\[\lambda =1\] in the center of the sphere, i.e. in equation (3).
\[\begin{align}
& \left( \dfrac{3-5\times 1}{2},1-2,1-2\times 1 \right) \\
& =\left( \dfrac{3-5}{2},-1,1-2 \right) \\
& =\left( -1,-1,-1 \right) \\
\end{align}\]
Hence, we got the center of the sphere as\[(-1,-1,-1)\].
Hence, option C is the correct answer.
Note: A greater circle or an orthodrome of a sphere is the intersection of the sphere and plane that passes through the center point of the sphere. Thus the intersection of a sphere and a plane is not empty or a single point, it’s a circle. Thus we consider \[S+\lambda L\].
Complete step-by-step answer:
Given the equation of circle as \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5=0\]
The equation of the plane is given as \[5x-2y+4z+7=0\].
In this question, we need to find the equation of the sphere.
The equation of the sphere is passing through the point of intersection of the sphere and the plane. Thus the equation of sphere can be said as \[\left( S+\lambda L \right)\],
Where S = equation of circle =\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5=0\]
L = equation of plane =\[5x-2y+4z+7=0\] and \[\lambda \]is constant.
\[S+\lambda L=\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5 \right)+\lambda \left( 5x-2y+4z+7 \right)=0\]
\[={{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( 3-5\lambda \right)x+\left( 4-2\lambda \right)y-\left( 2-4\lambda \right)z-\left( 5+7\lambda \right)=0......(1)\]
We know the general equation of sphere as,
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2wy+2vz+d=0.....(2)\]
Now let us compare equation (1) and equation (2).
From that we get,
\[\begin{align}
& 2u=-\left( 3-5\lambda \right)=5\lambda -3 \\
& 2w=\left( 4-2\lambda \right) \\
& 2v=-\left( 2-4\lambda \right)=4\lambda -2 \\
\end{align}\]
\[u=\dfrac{5\lambda -3}{2},w=\dfrac{4-2\lambda }{2}=2-\lambda ,v=\dfrac{4\lambda -2}{2}=2\lambda -1\]
Thus, the center of sphere is\[\left( -u,-v,-z \right)\]
\[=\left[ -\left( \dfrac{5\lambda -3}{2} \right),-\left( 2-\lambda \right),-\left( 2\lambda -1 \right) \right]\]
\[=\left[ \dfrac{3-5\lambda }{2},\lambda -2,1-2\lambda \right]......(3)\]
The center of the sphere lies on the plane if it is a great circle.
Therefore, let us substitute these values in the equation of the plane.
\[\begin{align}
& 5x-2y+4z+7=0 \\
& 5\left( \dfrac{3-5\lambda }{2} \right)-2\left( \lambda -2 \right)+4\left( 1-2\lambda \right)+7=0 \\
\end{align}\]
Open the brackets and simplify the expression to get the value of\[\lambda \].
\[=\dfrac{15}{2}-\dfrac{25\lambda }{2}-2\lambda +4+4-8\lambda +7=0\]
\[=\left( \dfrac{15}{2}+4+4+7 \right)-\lambda \left( \dfrac{25}{2}+2+8 \right)=0\]
\[=\left( \dfrac{15+8+8+14}{2} \right)-\lambda \left( \dfrac{25+4+16}{2} \right)=0\]
\[=\dfrac{45}{2}-\lambda \left( \dfrac{45}{2} \right)=0\]
\[\begin{align}
& =\dfrac{45-45\lambda }{2}=0 \\
& \therefore 45-45\lambda =0 \\
& \therefore 45=45\lambda \\
& \therefore \lambda =1 \\
\end{align}\]
Hence we got the value of constant\[\lambda =1\].
Now put the value of\[\lambda =1\] in equation (1).
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( 3-5\times 1 \right)x+\left( 4-2\times 1 \right)y-\left( 2-4\times 1 \right)z-\left( 5+7\times 1 \right)=0 \\
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( -2 \right)x+2y-\left( -2 \right)z-12=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2x+2y+2z-12=0.....(4) \\
\end{align}\]
Thus, we got the equation of the sphere.
Now, put the value of\[\lambda =1\] in the center of the sphere, i.e. in equation (3).
\[\begin{align}
& \left( \dfrac{3-5\times 1}{2},1-2,1-2\times 1 \right) \\
& =\left( \dfrac{3-5}{2},-1,1-2 \right) \\
& =\left( -1,-1,-1 \right) \\
\end{align}\]
Hence, we got the center of the sphere as\[(-1,-1,-1)\].
Hence, option C is the correct answer.
Note: A greater circle or an orthodrome of a sphere is the intersection of the sphere and plane that passes through the center point of the sphere. Thus the intersection of a sphere and a plane is not empty or a single point, it’s a circle. Thus we consider \[S+\lambda L\].
Recently Updated Pages
Identify the feminine gender noun from the given sentence class 10 english CBSE
Your club organized a blood donation camp in your city class 10 english CBSE
Choose the correct meaning of the idiomphrase from class 10 english CBSE
Identify the neuter gender noun from the given sentence class 10 english CBSE
Choose the word which best expresses the meaning of class 10 english CBSE
Choose the word which is closest to the opposite in class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
10 examples of friction in our daily life
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What is pollution? How many types of pollution? Define it