# The center of the sphere having the circle \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+4y-2z-5=0\],\[5x-2y+4z+7=0\] as the great circle is (section of a sphere by a plane through its center is known as a great circle)

A. \[\left( \dfrac{3}{2},-2,1 \right)\]

B. \[\left( 1,1,1 \right)\]

C. \[\left( -1,-1,-1 \right)\]

D. \[\left( 0,0,0 \right)\]

Last updated date: 28th Mar 2023

•

Total views: 306.6k

•

Views today: 7.83k

Answer

Verified

306.6k+ 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

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

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

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

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE