Question

The equation of the plane which cuts the sphere ${x^2} + {y^2} + {z^2} = {a^2}$ in a circle whose centre is $({{\alpha ,}}\,{{\beta ,}}\,{{\gamma }})$ is
A). $\alpha (x + \alpha ) + \beta (y + \beta ) + \gamma (z + \gamma ) = 0$
B). $\alpha (x + \alpha ) - \beta (y + \beta ) - \gamma (z + \gamma ) = 0$
C). \[{\alpha x + \beta y + \gamma z }={\alpha}^{2}+{\beta}^{2}+{\gamma }^{2}\]
D). ${\alpha}^{2}{x}+{\beta}^{2}{y}+{\gamma}^{2}z = \alpha + \beta + \gamma $

Answer 1

Hint: 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 ${x^2} + {y^2} + {z^2} = {a^2}$ is (0, 0, 0)
Also, we know that two vectors parallel then their direction ratios are proportional
 That is $\dfrac{{{a_1}}}{{{a_2}}}{\text{ = }}\dfrac{{{b_1}}}{{{b_2}}}{\text{ = }}\dfrac{{{c_1}}}{{{c_2}}}$
 The equation of the plane passing through the point \[({x_1},\,{y_1},\,{z_1})\,\] and perpendicular to the vector is\[{\text{ }}l(x{\text{ }} - {\text{ }}{x_1}) + \,m(y - {y_1}) + n\left( {z - {z_1}} \right) = 0\]

Complete step-by-step solution:
Given the equation of the sphere is ${x^2} + {y^2} + {z^2} = {a^2}$
We know that the center of the sphere ${x^2} + {y^2} + {z^2} = {a^2}$ is $O(0,0,0)$ and radius is a.
 Also given the center of the circle is A$({\alpha}\,{\beta}\,{\gamma})$.
The equation of the plane passing through the point \[({x_1},\,{y_1},\,{z_1})\,{\text{is}}\] and perpendicular to the vector is\[{\text{ l(x - }}{{\text{x}}_{\text{1}}}{\text{) + }}\,{\text{m(y - }}{{\text{y}}_{\text{1}}}{\text{) + n}}\left( {{\text{z - }}{{\text{z}}_{\text{1}}}} \right){\text{ = 0}}\]
 Here \[({x_1},\,{y_1},\,{z_1})\, = ({\alpha}\,{\beta}\,{\gamma})\]
Compare the corresponding coordinates then we get \[{x_1} = {{\alpha }},\,\,{y_1}\, = {{\beta }},\,\,{z_1}\, = \,\,{{\gamma }}\]
Then we get \[l(x - \alpha )+{{m(y -\beta)+}}n\left({z - \beta } \right)=0\]
  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 $\dfrac{{{a_1}}}{{{a_2}}}{\text{ = }}\dfrac{{{b_1}}}{{{b_2}}}{\text{ = }}\dfrac{{{c_1}}}{{{c_2}}}$
 Therefore, we get $\dfrac{{\text{l}}}{{{\alpha }}}{\text{ = }}\dfrac{{\text{m}}}{{{\beta }}}{\text{ = }}\dfrac{{\text{n}}}{{{\gamma }}}$
  By substituting the values of l, m, n we get
 ${{\alpha (x + \alpha ) - \beta (y + \beta ) - \gamma (z + \gamma ) = 0}}$
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 $r = \dfrac{d}{2}$ where d is the diameter and also the diameter is $d = 2r$