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Find two points on the z-axis and y-axis respectively, which are equidistant from the centreof the sphere x ^ 2 + y ^ 2 + z ^ 2 + 2x - 2y + 4z + 5 = 0

Question

Find two points on the z-axis and y-axis respectively, which are equidistant from the centre of the sphere

x2+y2+z2+2x2y+4z+5=0 x^2 + y^2 + z^2 + 2x - 2y + 4z + 5 = 0

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Solution

To find two points on the z-axis and y-axis that are equidistant from the center of the sphere, we need to first determine the center of the sphere.

The given equation of the sphere is x^2 + y^2 + z^2 + 2x - 2y + 4z + 5 = 0.

To find the center, we need to rewrite the equation in the standard form of a sphere, which is (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2, where (h, k, l) represents the center coordinates and r represents the radius.

Let's complete the square for the x, y, and z terms separately.

For the x terms: x^2 + 2x = (x^2 + 2x + 1) - 1 = (x + 1)^2 - 1

For the y terms: y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1

For the z terms: z^2 + 4z = (z^2 + 4z + 4) - 4 = (z + 2)^2 - 4

Now, let's substitute these expressions back into the original equation:

(x + 1)^2 - 1 + (y - 1)^2 - 1 + (z + 2)^2 - 4 + 5 = 0

Simplifying further:

(x + 1)^2 + (y - 1)^2 + (z + 2)^2 = 1

Comparing this equation with the standard form, we can see that the center of the sphere is (-1, 1, -2) and the radius is 1.

Now, to find two points on the z-axis and y-axis that are equidistant from the center, we can set the x-coordinate of these points to 0.

For the z-axis point, the coordinates will be (0, 0, z). To find the z-coordinate, we can substitute x = 0 and y = 0 into the equation of the sphere:

(0 + 1)^2 + (0 - 1)^2 + (z + 2)^2 = 1

1 + 1 + (z + 2)^2 = 1

(z + 2)^2 = -1

Since the square of a real number cannot be negative, there are no points on the z-axis that are equidistant from the center.

For the y-axis point, the coordinates will be (0, y, 0). To find the y-coordinate, we can substitute x = 0 and z = 0 into the equation of the sphere:

(0 + 1)^2 + (y - 1)^2 + (0 + 2)^2 = 1

1 + (y - 1)^2 + 4 = 1

(y - 1)^2 + 4 = 0

Again, since the square of a real number cannot be negative, there are no points on the y-axis that are equidistant from the center.

Therefore, there are no points on the z-axis and y-axis that are equidistant from the center of the given sphere.

This problem has been solved

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