Sphere equation

Obtain the equation of sphere its centre on the line 3x+2z=0=4x-5y and passes through the points (0,-2,-4) and (2,-1,1).

Correct result:

r =  3.9398
x₀ =  1.4894
y₀ =  1.1915
z₀ =  -2.234

Solution:

$$\begin{gathered} 3x+2z=0 \\ 4x-5y=0 \\ {r}^2=(x-0{)}^2+(y+2{)}^2+(z+4{)}^2 \\ {r}^2=(x-2{)}^2+(y+1{)}^2+(z-1{)}^2 \\ (x-x_0{)}^2+(y-y_0{)}^2+(z-z_0{)}^2=r^2 \\ r>0 \\ r=\sqrt{34289}\mathrm{/}47=3.9398 \end{gathered}$$

$x_0=70\mathrm{/}47={\displaystyle \frac{70}{47}}=1.4894$

$y_0=56\mathrm{/}47={\displaystyle \frac{56}{47}}=1.1915$

$z_{0}=-105/47=- \dfrac{ 105 }{ 47 }=-2.234$

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