Question

Show that the points (3, 2, 2), (–1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.

Answer 1

Let the points be A (3, 2, 2), B ($-$1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere
whose centre be P (1, 3, 4).
Since, AP, BP, CP and DP are radii.
Hence, AP = BP = CP = DP

$$\begin{gathered} AP=\sqrt{{\left(1-3\right)}^2+{\left(3-2\right)}^2+{\left(4-2\right)}^2} \\ =\sqrt{{\left(-2\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2} \\ =\sqrt{4+1+4} \\ =\sqrt{9} \\ =3 \\ BP=\sqrt{{\left(1+1\right)}^2+{\left(3-4\right)}^2+{\left(4-2\right)}^2} \\ =\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(2\right)}^2} \\ =\sqrt{4+1+4} \\ =\sqrt{9} \\ =3 \\ CP=\sqrt{{\left(1-0\right)}^2+{\left(3-5\right)}^2+{\left(4-6\right)}^2} \\ =\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{1+4+4} \\ =\sqrt{9} \\ =3 \\ DP=\sqrt{{\left(1-2\right)}^2+{\left(3-1\right)}^2+{\left(4-2\right)}^2} \\ =\sqrt{{\left(-1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2} \\ =\sqrt{1+4+4} \\ =\sqrt{9} \\ =3 \end{gathered}$$

Here, we see that AP = BP = CP = DP
Hence, A (3, 2, 2), B ($-$1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere whose radius is 3 .