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# 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. $\left( 0,0,0 \right)$  Answer Verified
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 $S+\lambda L$.
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