Chapter 10, Problem 4CT

To determine

Tofind:The equation of sphere whose end points of diameter are given. Also, to sketch the sphere and the xz- plane.

Answer to Problem 4CT

The equation of sphere is ${\left(x-7\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2=19$ . The graph of sphere and xz-plane is shown in Figure-2.

Explanation of Solution

Given information:

The end points of the diameter of the sphere are $A\left(8,-2,5\right)$ and $B\left(6,4,-1\right)$

Formulaused:

The midpoint of diameter is the center of the sphere. The coordinates of mid-point of line segment joining points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ are $x=\frac{x_1+x_2}{2}$ , $y=\frac{y_1+y_2}{2}$ , and $z=\frac{z_1+z_2}{2}$ ,

Calculations:

The end points of diameter of the sphere are $A\left(8,-2,5\right)$ , and $B\left(6,4,-1\right)$ . The midpoint of the diameter is the center $\left(x_C,y_C,z_C\right)$ of the sphere given by

  

  $x_C=\frac{8+6}{2}=7$ , $y_C=\frac{-2+4}{2}=1$ , $z_C=\frac{5-1}{2}=2$ .

The radius Rof the sphereis $PC=AC=CB$ .

The radius of the sphere $R=AC=\frac{1}{2}AB=\frac{1}{2}\sqrt{{\left(6-8\right)}^2+{\left(4-(-2)\right)}^2+{\left(-1-5\right)}^2}$

  $\Rightarrow R=AC=\frac{1}{2}AB=\frac{1}{2}\sqrt{4+36+36}$

  $\Rightarrow R=\frac{1}{2}\sqrt{76}=\frac{1}{2}2\sqrt{19}=\sqrt{19}$

Let $P\left(x,y,z\right)$ be a point on the sphere then $PA=R$

  $\sqrt{{\left(x-7\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2}=\sqrt{19}$

  $\Rightarrow {\left(x-7\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2=19$

Thus, the equation of sphere is ${\left(x-7\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2=19$ .

The diagram of sphere is shown in Figure (b) here.

  

Conclusion:

The equation of sphere is ${\left(x-7\right)}^2+{\left(y-1\right)}^2+{\left(z-2\right)}^2=19$ . The graph of sphere and xz-plane is shown in Figure-2.