Question

A sphere of constant radius k passes through the origin and meets axes in A, B, and C. The centroid of the triangle ABC lies on the sphere
(A) $9\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=4{{k}^{2}}$
(B) $3\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=4{{k}^{2}}$
(C) $\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=4{{k}^{2}}$
(D) None of these

Answer 1

Assume that the equation of the sphere whose coordinate of the center is $\left( a,b,c \right)$ and radius is equal to k is ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}+{{\left( z-c \right)}^{2}}={{k}^{2}}$ . Since it is given that the sphere is passing through the origin so, the equation of the sphere must satisfy the coordinates of the origin. Put $x=0,y=0,\,\text{and }z=0$ in the equation of the sphere and simplify it further to get an equation in terms of $a,b$ , and $c$ . It is also given that the sphere meets the axes at points A, B, and C respectively. Since the sphere meets the x-axis at point A so, point A must have the y and z coordinates of point A equal to zero. Now, put, $y=0$ and $z=0$ in the equation of the sphere and replace ${{k}^{2}}$ by $\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)$ . Then, get the value of x which is x-coordinate of point A. Since the sphere meets the y-axis at point B so, point B must have the x and z coordinates of point B equal to zero. Now, put $x=0$ and $y=0$ in the equation of the sphere and replace ${{k}^{2}}$ by $\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)$ . Then, get the value of y which is the y-coordinate of point B. Since the sphere meets the z-axis at point C so, point C must have the x and y coordinates of point C equal to zero. Now, put $x=0$ and $y=0$ in the equation of the sphere and replace ${{k}^{2}}$ by $\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)$ . Then, get the value of z which is z-coordinate of point C. Now, use the formula $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$ and get the coordinate of the centroid of $\Delta ABC$ . Now, square the x-coordinate, y-coordinate, and the z-coordinate of the centroid of $\Delta ABC$ and then add them. At last, replace $\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)$ by ${{k}^{2}}$ and get the equation of the centroid of $\Delta ABC$ .

Complete step by step answer:
First of all, let us assume the equation of the sphere whose radius is equal to k be
${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}+{{\left( z-c \right)}^{2}}={{k}^{2}}$ …………………………….(1)
Since the sphere is passing through the origin so, the equation of the sphere must satisfy the coordinates of the origin.
The coordinate of the origin is $\left( 0,0,0 \right)$ .
Now, putting, $x=0,y=0,\,\text{and }z=0$ in equation (1), we get
$\Rightarrow {{\left( 0-a \right)}^{2}}+{{\left( 0-b \right)}^{2}}+{{\left( 0-c \right)}^{2}}={{k}^{2}}$
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}={{k}^{2}}$ …………………………………..(2)
We know that when this sphere intersects the x-axis at a point then, y coordinate and z coordinate of that point becomes zero.
Now, putting $y=0$ and $z=0$ in equation (1), we get
$\Rightarrow {{\left( x-a \right)}^{2}}+{{\left( 0-b \right)}^{2}}+{{\left( 0-c \right)}^{2}}={{k}^{2}}$
$\Rightarrow {{\left( x-a \right)}^{2}}+{{b}^{2}}+{{c}^{2}}={{k}^{2}}$ …………………………………(3)
From equation (2) and equation (3)

& \Rightarrow {{\left( x-a \right)}^{2}}+{{b}^{2}}+{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\\ & \Rightarrow {{\left( x-a \right)}^{2}}={{a}^{2}} \\\ \end{aligned}$$ $$\Rightarrow x=a+a$$ $$\Rightarrow x=2a$$ ……………………..(4) So, the x coordinate of the point where the sphere intersects the x-axis is 2a. The coordinate of the point of intersection of the sphere and the x-axis is A $$\left( 2a,0,0 \right)$$ …………………………..(5) We know that when this sphere intersects the y-axis at a point then, x coordinate and z coordinate of that point becomes zero. Now, putting $$x=0$$ and $$z=0$$ in equation (1), we get $$\Rightarrow {{\left( 0-a \right)}^{2}}+{{\left( y-b \right)}^{2}}+{{\left( 0-c \right)}^{2}}={{k}^{2}}$$ $$\Rightarrow {{a}^{2}}+{{\left( y-b \right)}^{2}}+{{c}^{2}}={{k}^{2}}$$ …………………………………(6) From equation (2) and equation (6) $$\begin{aligned} & \Rightarrow {{a}^{2}}+{{\left( y-b \right)}^{2}}+{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\\ & \Rightarrow {{\left( y-b \right)}^{2}}={{b}^{2}} \\\ \end{aligned}$$ $$\Rightarrow y=b+b$$ $$\Rightarrow y=2b$$ ……………………..(7) So, the y coordinate of the point where the sphere intersects the y-axis is 2b. The coordinate of the point of intersection of the sphere and the y-axis is B $$\left( 0,2b,0 \right)$$ …………………………..(8) We know that when this sphere intersects the z-axis at a point then, x coordinate and y coordinate of that point becomes zero. Now, putting $$x=0$$ and $$y=0$$ in equation (1), we get $$\Rightarrow {{\left( 0-a \right)}^{2}}+{{\left( 0-b \right)}^{2}}+{{\left( z-c \right)}^{2}}={{k}^{2}}$$ $$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{\left( z-c \right)}^{2}}={{k}^{2}}$$ …………………………………(9) From equation (2) and equation (9) $$\begin{aligned} & \Rightarrow {{a}^{2}}+{{b}^{2}}+{{\left( z-c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\\ & \Rightarrow {{\left( z-c \right)}^{2}}={{c}^{2}} \\\ \end{aligned}$$ $$\Rightarrow z=c+c$$ $$\Rightarrow z=2c$$ ……………………..(10) So, the z coordinate of the point where the sphere intersects the y-axis is 2c. The coordinate of the point of intersection of the sphere and the z-axis is C $$\left( 0,0,2c \right)$$ …………………………..(11) Now, we have $$\Delta ABC$$ . We need to find the centroid of $$\Delta ABC$$ . We know the formula that the coordinates of the centroid of a triangle is $$\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$ ……………………………..(13) Now, from equation (10), equation (11), and equation (12), we have the coordinates of the points where the sphere intersects the x-axis, y-axis, and the z-axis ie, A, B, and C. The coordinate of point A = $$\left( 2a,0,0 \right)$$ The coordinate of point B = $$\left( 0,2b,0 \right)$$ . The coordinate of point C = $$\left( 0,0,2c \right)$$ . The coordinate of the centroid of $$\Delta ABC$$ = $$\left( \dfrac{2a+0+0}{3},\dfrac{0+2b+0}{3},\dfrac{0+0+2c}{3} \right)=\left( \dfrac{2a}{3},\dfrac{2b}{3},\dfrac{2c}{3} \right)$$ ………………………………….(14) The x coordinate of the centroid, x = $$\dfrac{2a}{3}$$ ……………………..……..(15) On squaring we get, $${{x}^{2}}=\dfrac{4{{a}^{2}}}{9}$$ ……………………………………(16) The y coordinate of the centroid, y = $$\dfrac{2b}{3}$$ ………………………....(17) On squaring we get, $${{y}^{2}}=\dfrac{4{{b}^{2}}}{9}$$ ……………………………………(18) The z coordinate of the centroid, z = $$\dfrac{2c}{3}$$ …………………..(19) On squaring we get, $${{z}^{2}}=\dfrac{4{{c}^{2}}}{9}$$ ……………………………………(20) Now, adding equation (16), equation (18), and equation (20), we get $$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=\dfrac{4{{a}^{2}}}{9}+\dfrac{4{{b}^{2}}}{9}+\dfrac{4{{c}^{2}}}{9}$$ $$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=\dfrac{4\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}{9}$$ …………………………….(21) Now, from equation (2) and equation (21), we get $$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=\dfrac{4{{k}^{2}}}{9}$$ $$\Rightarrow 9\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=4{{k}^{2}}$$ ………………………..(22) Therefore, the equation of the sphere where the centroid of $$\Delta ABC$$ lies is $$9\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=4{{k}^{2}}$$ . **So, the correct answer is “Option A”.** **Note:** In this question, one might think to assume the equation of the sphere equal to $$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)={{k}^{2}}$$ . This is wrong. Since it is given that the sphere is passing through the origin and this equation doesn’t satisfy the coordinates of origin. Let us take the equation of the sphere $$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)={{k}^{2}}$$ , where k is the radius. The sphere is passing through the origin. So, the equation of the sphere must satisfy the coordinate of origin. Now, putting, $$x=0,y=0,\,\text{and }z=0$$ in the equation $$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)={{k}^{2}}$$ , we get $$\begin{aligned} & \Rightarrow \left( {{0}^{2}}+{{0}^{2}}+{{0}^{2}} \right)={{k}^{2}} \\\ & \Rightarrow 0={{k}^{2}} \\\ & \Rightarrow k=0 \\\ \end{aligned}$$ The value of k becomes equal to zero. But k is the radius of a sphere and the value of the radius of a sphere cannot be equal to zero. Therefore, it is a contradiction. Hence, we cannot take the equation of the sphere as $$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)={{k}^{2}}$$ .