Question
If the sum of the squares of the distance of the point $$\left( {x,\,y,\,z} \right)$$ from the points $$\left( {a,\,0,\,0} \right)$$ and $$\left( { - a,\,0,\,0} \right)$$ is $$2{c^2},$$ then which one of the following is correct ?
A.
$${x^2} + {a^2} = 2{c^2} - {y^2} - {z^2}$$
B.
$${x^2} + {a^2} = {c^2} - {y^2} - {z^2}$$
C.
$${x^2} - {a^2} = {c^2} - {y^2} - {z^2}$$
D.
$${x^2} + {a^2} = {c^2} + {y^2} + {z^2}$$
Answer :
$${x^2} + {a^2} = {c^2} - {y^2} - {z^2}$$
Solution :
$$\eqalign{
& {\text{Let the point be }}P\left( {x,\,y,\,z} \right){\text{ and two points,}} \cr
& \left( {a,\,0,\,0} \right){\text{ and }}\left( { - a,\,0,\,0} \right){\text{ be }}A{\text{ and }}B \cr
& {\text{As given in the problem,}}\,P{A^2} + P{B^2} = 2{c^2} \cr
& {\text{So,}} \cr
& {\left( {x + a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {z - 0} \right)^2} + {\left( {x - a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {z - 0} \right)^2} = 2{c^2} \cr
& {\text{or,}}\,{\left( {x + a} \right)^2} + {y^2} + {z^2} + {\left( {x - a} \right)^2} + {y^2} + {z^2} = 2{c^2} \cr
& \Rightarrow {x^2} + 2a + {a^2} + {y^2} + {z^2} + {x^2} - 2a + {a^2} + {y^2} + {z^2} = 2{c^2} \cr
& \Rightarrow 2\left( {{x^2} + {y^2} + {z^2} + {a^2}} \right) = 2{c^2} \cr
& \Rightarrow {x^2} + {y^2} + {z^2} + {a^2} = {c^2} \cr
& \Rightarrow {x^2} + {a^2} = {c^2} - {y^2} - {z^2} \cr} $$