Question

Find the equation of the sphere passing through the points (3, 0, 0), (0, -1, 0) and (0, 0, -2) and having the center of the plane $3x+2y+4z=1$.

Answer 1

Hint: Substitute the given points in the general equation of a sphere. Thus get 3 equations, take the center as (h, k, l). Find the value of (h, k, l) and substitute it back in one of the equations of the sphere we found. Simplify it to get the equation of sphere.

Complete step-by-step answer:
We have been given three points through which a sphere passes. The three points are (3, 0, 0), (0, -1, 0) and (0, 0, -2).
The general equation of a sphere is given by the formula,
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}+{{\left( z-l \right)}^{2}}={{r}^{2}}-(1)$, where (h, k, l) represents the center of the sphere and r is the radius of the sphere.
Now we need to find the equation of the sphere connecting the three points (3, 0, 0), (0, -1, 0) and (0, 0, -2).
Thus let us substitute (3, 0, 0) in equation (1).

& {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}+{{\left( z-l \right)}^{2}}={{r}^{2}} \\\ & {{\left( 3-h \right)}^{2}}+{{\left( 0-k \right)}^{2}}+{{\left( 0-l \right)}^{2}}={{r}^{2}} \\\ & \Rightarrow {{\left( 3-h \right)}^{2}}+{{k}^{2}}+{{l}^{2}}={{r}^{2}}-(2) \\\ \end{aligned}$$ Now consider (0, -1, 0) and (x, y, z) and put in equation (1). $$\begin{aligned} & {{\left( 0-h \right)}^{2}}+{{\left( 1+k \right)}^{2}}+{{\left( 0-l \right)}^{2}}={{r}^{2}} \\\ & \Rightarrow {{h}^{2}}+{{\left( 1+k \right)}^{2}}+{{l}^{2}}={{r}^{2}}-(3) \\\ \end{aligned}$$ Now take the points (0, 0, -2) and (x, y, z) and put in equation (1). $$\begin{aligned} & {{\left( 0-h \right)}^{2}}+{{\left( 0-k \right)}^{2}}+{{\left( -2+l \right)}^{2}}={{r}^{2}} \\\ & {{h}^{2}}+{{k}^{2}}+{{\left( 2+l \right)}^{2}}={{r}^{2}}-(4) \\\ \end{aligned}$$ Now we have been given an equation of the center of the plane, $$3x+2y+4z=1$$. $$\begin{aligned} & {{\left( 3-h \right)}^{2}}+{{k}^{2}}+{{l}^{2}}={{r}^{2}}-(2) \\\ & {{h}^{2}}+{{\left( 1+k \right)}^{2}}+{{l}^{2}}={{r}^{2}}-(3) \\\ & {{h}^{2}}+{{k}^{2}}+{{\left( 2+l \right)}^{2}}={{r}^{2}}-(4) \\\ & 3x+2y+4z=1-(5) \\\ \end{aligned}$$ Now let us equate (2) and (3). $${{\left( 3-h \right)}^{2}}+{{k}^{2}}+{{l}^{2}}={{h}^{2}}+{{\left( k+1 \right)}^{2}}+{{l}^{2}}$$ Cancel out the like terms, $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$. $${{h}^{2}}+9-6h+{{k}^{2}}+{{l}^{2}}={{h}^{2}}+{{k}^{2}}+2k+1+{{l}^{2}}$$ Cancel out $${{x}^{2}},{{y}^{2}}$$ and $${{z}^{2}}$$. Thus, we get, $$\begin{aligned} & 9-6h=2k+1 \\\ & \Rightarrow 6h+2k-8=0\Rightarrow 3h+k=4-(6) \\\ \end{aligned}$$ Similarly, let us equate (3) and (4) and simplify it, $$\begin{aligned} & {{h}^{2}}+{{\left( k+1 \right)}^{2}}+{{l}^{2}}={{h}^{2}}+{{k}^{2}}+{{\left( l+2 \right)}^{2}} \\\ & {{h}^{2}}+{{k}^{2}}+2k+1+{{l}^{2}}={{h}^{2}}+{{k}^{2}}+{{l}^{2}}+4l+4 \\\ \end{aligned}$$ Cancel out $${{x}^{2}},{{y}^{2}}$$ and $${{z}^{2}}$$ from LHS and RHS. $$\begin{aligned} & 2k+1=4l+4 \\\ & \Rightarrow 2k-4l=3-(7) \\\ \end{aligned}$$ Now let us equate (2) and (4). $$\begin{aligned} & {{\left( {{3}^{2}}-h \right)}^{2}}+{{k}^{2}}+{{l}^{2}}={{h}^{2}}+{{k}^{2}}+{{\left( l+1 \right)}^{2}} \\\ & {{h}^{2}}-6h+9+{{k}^{2}}+{{l}^{2}}={{h}^{2}}+{{k}^{2}}+{{l}^{2}}+4l+4 \\\ & -6h+9=4l+4 \\\ & \Rightarrow 6h+4l-5=0 \\\ \end{aligned}$$ i.e. $$6h+4l=5-(8)$$ Now, in equation (6), $$3h=4-k$$. In equation (7), $$2k=3+4l$$ In equation (7), $$4l=2k-3$$ We have been given that, $$3x+2y+4z=1$$, now let us substitute the above values in this equation, $$3x+2y+4z=1$$, write h and l as x and z in equation (6) and (7) as the centre (h,k,l) passes through the given equation we get $$3x=4-k$$ and $$4z=2k-3$$. $$\begin{aligned} & \therefore \left( 4-k \right)+2k+\left( 2k-3 \right)=1 \\\ & 4-k+2k+2k-3=1 \\\ \end{aligned}$$ $$3k+1=1$$$$\Rightarrow 3k=0$$ and $$k=0$$. Now, put $$k=0$$ in $$4l=2k-3$$. $$\begin{aligned} & 4l=2\times 0-3 \\\ & \therefore l=\dfrac{-3}{4} \\\ \end{aligned}$$ Now, put $$k=0$$ in $$3h=4-k$$ $$\therefore 3h=4-0$$ i.e. $$h=\dfrac{4}{3}$$ Hence, we got, $$h=\dfrac{4}{3}$$, $$k=0$$ and $$l=\dfrac{-3}{4}$$. In equation (2), put these values, $$\begin{aligned} & {{\left( 3-h \right)}^{2}}+{{k}^{2}}+{{l}^{2}}={{r}^{2}} \\\ & {{\left( 3-\dfrac{4}{3} \right)}^{2}}+{{0}^{2}}+{{\left( \dfrac{-3}{4} \right)}^{2}}={{r}^{2}}\Rightarrow {{\left[ \dfrac{9-5}{3} \right]}^{2}}+\dfrac{9}{16}={{r}^{2}} \\\ \end{aligned}$$ $$r=\sqrt{\dfrac{25}{9}+\dfrac{9}{16}}=\sqrt{\dfrac{481}{144}}$$ or $${{r}^{2}}=\dfrac{481}{144}$$. Thus, we can write the equation of sphere is, $${{\left( x-\dfrac{4}{3} \right)}^{2}}+{{y}^{2}}+{{\left( z+\dfrac{3}{4} \right)}^{2}}=\dfrac{481}{144}$$ i.e. general equation of sphere $$={{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}+{{\left( z-l \right)}^{2}}={{r}^{2}}$$ We got, $$\left( x,y,z \right)=\left( \dfrac{4}{3},0,\dfrac{-3}{4} \right)=\left( h,k,l \right)$$. Thus substituting these values we get the equation of sphere. We can simplify the equation of sphere as, $$\begin{aligned} & \dfrac{{{\left( 3x-4 \right)}^{2}}}{{{3}^{2}}}+{{y}^{2}}+\dfrac{{{\left( 4z+3 \right)}^{2}}}{{{4}^{2}}}=\dfrac{481}{144} \\\ & \dfrac{{{\left( 3x-4 \right)}^{2}}}{9}+{{y}^{2}}+\dfrac{{{\left( 4z+3 \right)}^{2}}}{16}=\dfrac{481}{144} \\\ & \dfrac{16{{\left( 3x-4 \right)}^{2}}+144{{y}^{2}}+9{{\left( 4z+3 \right)}^{2}}}{144}=\dfrac{481}{144} \\\ \end{aligned}$$ Cancel out 144 from denominators of LHS and RHS. Thus we get, $$\begin{aligned} & 16{{\left( 3x-4 \right)}^{2}}+144{{y}^{2}}+9{{\left( 4z+3 \right)}^{2}}=481 \\\ & 16\left[ {{\left( 3x \right)}^{2}}-2\times 3x\times 4+16 \right]+144{{y}^{2}}+9\left[ {{\left( 4z \right)}^{2}}+2\times 4z\times 3+9 \right]=481 \\\ & 16\left[ 9{{x}^{2}}-24x+16 \right]+144{{y}^{2}}+9\left[ 16{{z}^{2}}+24z+9 \right]=481 \\\ & 144{{x}^{2}}-384x+256+144{{y}^{2}}+144{{z}^{2}}+216z+81=481 \\\ & 144{{x}^{2}}+144{{y}^{2}}+144{{z}^{2}}-384x+216z=481-256-81 \\\ \end{aligned}$$ i.e. $$144\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)-384x+216z=144$$ Now let us divide the entire equation by 24. $$\begin{aligned} & 6\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)-16x+9z=6 \\\ & \Rightarrow 6{{x}^{2}}+6{{y}^{2}}+6{{z}^{2}}-16x+9z-6=0 \\\ \end{aligned}$$ Hence, we got the equation of sphere as, $$6{{x}^{2}}+6{{y}^{2}}+6{{z}^{2}}-16x+9z-6=0$$ Note: The equation of a sphere is similar to that of a circle, but with an extra variable for the extra dimension. $${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$$ is the general equation of a circle, where as the general equation of sphere is $${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}} + {{\left( z-l \right)}^{2}}={{r}^{2}}$$. The coordinate (h, k, l) tells us the center of the sphere.