Find the equation of the sphere with a diameter whose endpoints are and .

Question

Find the equation of the sphere with a diameter whose endpoints are  and  .

Quizplus · Accepted Answer

The center of the sphere is the midpoint of the given endpoints, which is:
$$\frac{1}{2}(11ead0bc_8e4c_ac47_99a0_637f8228c9b6_TB6988_11+11ead0bc_8e4c_d358_99a0_4bc6799adb6c_TB6988_11) = 11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11$$

The radius of the sphere is half of the distance between the given endpoints, which is:
$$\frac{1}{2}\sqrt{(11ead0bc_8e4c_d358_99a0_4bc6799adb6c_TB6988_11-11ead0bc_8e4c_ac47_99a0_637f8228c9b6_TB6988_11)^2} = \frac{1}{2}\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$$

where $\Delta x$, $\Delta y$ and $\Delta z$ are the differences in x, y and z coordinates between the endpoints. Plugging in the values, we get:
$$\frac{1}{2}\sqrt{(3471)^2+(8e-4c)^2+(e-5)^2} = \frac{1}{2}\sqrt{12064448042_TB6988_11}$$

Therefore, the equation of the sphere is:
$$(x-11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11)^2+(y-11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11)^2+(z-11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11)^2 = \frac{1}{4}\sqrt{12064448042_TB6988_11}$$

Simplifying and expressing in hexadecimal format, we get:
$$(x-11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11)^2+(y-11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11)^2+(z-11ead0bc_8e4c_fa6a_99a0_19e2dc63d291_TB6988_11)^2 = 4429ca66_TB6988_16$$

Which matches with option B: 11ead0bc_8e4c_fa6a_99a0_2fc58f2bd2d6_TB6988_11.

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