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Question: The equation of the plane which cuts the sphere \({x^2} + {y^2} + {z^2} = {a^2}\) in a circle whose ...

The equation of the plane which cuts the sphere x2+y2+z2=a2{x^2} + {y^2} + {z^2} = {a^2} in a circle whose centre is (α, β, γ)({{\alpha ,}}\,{{\beta ,}}\,{{\gamma }}) is
A). α(x+α)+β(y+β)+γ(z+γ)=0\alpha (x + \alpha ) + \beta (y + \beta ) + \gamma (z + \gamma ) = 0
B). α(x+α)−β(y+β)−γ(z+γ)=0\alpha (x + \alpha ) - \beta (y + \beta ) - \gamma (z + \gamma ) = 0
C). αx+βy+γz=α2+β2+γ2{\alpha x + \beta y + \gamma z }={\alpha}^{2}+{\beta}^{2}+{\gamma }^{2}
D). α2x+β2y+γ2z=α+β+γ{\alpha}^{2}{x}+{\beta}^{2}{y}+{\gamma}^{2}z = \alpha + \beta + \gamma

Explanation

Solution

Since from the given that the equations of the plane and its center is given. Where radius means half of the overall diameter length.
We know that the center of the sphere x2+y2+z2=a2{x^2} + {y^2} + {z^2} = {a^2} is (0, 0, 0)
Also, we know that two vectors parallel then their direction ratios are proportional
That is a1a2 = b1b2 = c1c2\dfrac{{{a_1}}}{{{a_2}}}{\text{ = }}\dfrac{{{b_1}}}{{{b_2}}}{\text{ = }}\dfrac{{{c_1}}}{{{c_2}}}
The equation of the plane passing through the point (x1, y1, z1) ({x_1},\,{y_1},\,{z_1})\, and perpendicular to the vector is l(x − x1)+ m(y−y1)+n(z−z1)=0{\text{ }}l(x{\text{ }} - {\text{ }}{x_1}) + \,m(y - {y_1}) + n\left( {z - {z_1}} \right) = 0

Complete step-by-step solution:
Given the equation of the sphere is x2+y2+z2=a2{x^2} + {y^2} + {z^2} = {a^2}
We know that the center of the sphere x2+y2+z2=a2{x^2} + {y^2} + {z^2} = {a^2} is O(0,0,0)O(0,0,0) and radius is a.
Also given the center of the circle is A(α β γ)({\alpha}\,{\beta}\,{\gamma}).
The equation of the plane passing through the point (x1, y1, z1) is({x_1},\,{y_1},\,{z_1})\,{\text{is}} and perpendicular to the vector is l(x - x1) +  m(y - y1) + n(z - z1) = 0{\text{ l(x - }}{{\text{x}}_{\text{1}}}{\text{) + }}\,{\text{m(y - }}{{\text{y}}_{\text{1}}}{\text{) + n}}\left( {{\text{z - }}{{\text{z}}_{\text{1}}}} \right){\text{ = 0}}
Here (x1, y1, z1) =(α β γ)({x_1},\,{y_1},\,{z_1})\, = ({\alpha}\,{\beta}\,{\gamma})
Compare the corresponding coordinates then we get x1=α,  y1 =β,  z1 =  γ{x_1} = {{\alpha }},\,\,{y_1}\, = {{\beta }},\,\,{z_1}\, = \,\,{{\gamma }}
Then we get l(x−α)+m(y−β)+n(z−β)=0l(x - \alpha )+{{m(y -\beta)+}}n\left({z - \beta } \right)=0
Since OA is perpendicular to the plane
Therefore normal is parallel to OA
We know that two vectors parallel then their direction ratios are proportional
That is a1a2 = b1b2 = c1c2\dfrac{{{a_1}}}{{{a_2}}}{\text{ = }}\dfrac{{{b_1}}}{{{b_2}}}{\text{ = }}\dfrac{{{c_1}}}{{{c_2}}}
Therefore, we get lα = mβ = nγ\dfrac{{\text{l}}}{{{\alpha }}}{\text{ = }}\dfrac{{\text{m}}}{{{\beta }}}{\text{ = }}\dfrac{{\text{n}}}{{{\gamma }}}
By substituting the values of l, m, n we get
α(x+α)−β(y+β)−γ(z+γ)=0{{\alpha (x + \alpha ) - \beta (y + \beta ) - \gamma (z + \gamma ) = 0}}
Option (B) is correct.

Note: Thus, using the given information, we simply solved the above problem and If the particular value of the radius of the sphere and center of the circle is given then we get the particular equation of the plane. If the equation of the sphere contains a center other than the origin then the resultant equation changes.
Since radius can be framed as r=d2r = \dfrac{d}{2} where d is the diameter and also the diameter is d=2rd = 2r