Question

For constants \(a, b, c,\) and \(d,\) show that the equation $$x^{2}+y^{2}+z^{2}-2 a x-2 b y-2 c z=d$$ describes a sphere centered at \((a, b, c)\) with radius \(r,\) where \(r^{2}=d+a^{2}+b^{2}+c^{2},\) provided \(d+a^{2}+b^{2}+c^{2}>0\)

Short answer

If so, what is its center and radius? Answer: Yes, the given equation represents a sphere when \(d + a^{2} + b^{2} + c^{2} > 0\). The sphere has a center at \((a, b, c)\) and a radius of \(r = \sqrt{d + a^{2} + b^{2} + c^{2}}\).

Step-by-step solution

Rewrite the equation

The given equation is: $$x^{2}+y^{2}+z^{2}-2a x-2b y-2c z=d$$ We can rewrite this equation by rearranging its terms: $$(x^{2}-2ax) +(y^{2}-2by)+(z^{2}-2cz)=d$$

Complete the square

Now, we will complete the square for each variable, i.e. completing the square for \(x\), \(y\), and \(z\). To complete the square, we will add and subtract the square of half of the coefficients of the linear terms. So, for \(x\), since the linear term is \(-2ax\), we will add and subtract \((a)^2\). Similarly for \(y\) and \(z\), we will add and subtract \((b)^2\) and \((c)^2\), respectively: $$(x^{2}-2ax + a^{2}) +(y^{2}-2by + b^{2})+(z^{2}-2cz + c^{2})=d + a^2 + b^2 + c^2$$

Rewrite the equation as a sum of squares

Now we can rewrite the above equation as a sum of squares for each variable: $$(x - a)^{2} +(y - b)^{2} + (z - c)^{2} = d + a^{2} + b^{2} + c^{2}$$

Identify the center and radius

Recall that the equation of a sphere with center \((a, b, c)\) and radius \(r\) can be written as: $$(x - a)^{2} + (y - b)^{2} + (z - c)^{2} = r^2$$ Comparing this equation with the equation derived in Step 3, we can see that the given equation indeed represents a sphere centered at \((a, b, c)\) with radius \(r\), where \(r^{2} = d + a^{2} + b^{2} + c^{2}\) provided \(d + a^{2} + b^{2} + c^{2} > 0\).