The equation of sphere is x2 + y2 + z2 - x + z - 2 = 0, its radius is
The given equation of a sphere is \(x^2 + y^2 + z^2 - x + z - 2 = 0\). We need to find its radius. To do this, we compare the given equation with the standard general form of the equation of a sphere.
The general equation of a sphere in three dimensions is given by:
\(x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0\)
In this standard form, the center of the sphere is at the point \((-u, -v, -w)\) and the radius of the sphere is given by the formula:
\(r = \sqrt{u^2 + v^2 + w^2 - d}\)
Let's compare the given equation \(x^2 + y^2 + z^2 - x + z - 2 = 0\) with the general form \(x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0\).
By comparing the coefficients of \(x, y, z\), and the constant term, we can find the values of \(u, v, w\), and \(d\).
Now that we have the values of \(u, v, w\), and \(d\), we can use the formula for the radius of the sphere, \(r = \sqrt{u^2 + v^2 + w^2 - d}\), to calculate the radius of the given sphere.
Substitute the values we found:
\(u = -\frac{1}{2}\), \(v = 0\), \(w = \frac{1}{2}\), \(d = -2\)
\(r = \sqrt{\left(-\frac{1}{2}\right)^2 + (0)^2 + \left(\frac{1}{2}\right)^2 - (-2)}\)
\(r = \sqrt{\frac{1}{4} + 0 + \frac{1}{4} + 2}\)
Combine the fractions:
\(r = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{8}{4}}\)
\(r = \sqrt{\frac{1+1+8}{4}}\)
\(r = \sqrt{\frac{10}{4}}\)
Simplify the fraction inside the square root:
\(r = \sqrt{\frac{5}{2}}\)
Thus, the radius of the sphere is \(\sqrt{\frac{5}{2}}\).
To find the radius of a sphere from its general equation \(x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0\):
Applying this method to the given equation \(x^2 + y^2 + z^2 - x + z - 2 = 0\), we successfully calculated the radius of sphere as \(\sqrt{\frac{5}{2}}\).
This confirms the process for determining the radius of sphere from its algebraic form.
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