Answer to Question #332777 in Analytic Geometry for ajee

Question

Find the equation (formula) of a sphere with radius r and center C(h, k, l) and show that x² + y² + z² - 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find its center and radius

Answer 1

$x^2 + y^2 + z^2 - 6x + 2y + 8z - 4 = 0 \\ x^2-6x+9-9+y^2+2y+1-1+z^2+8z+16-16-4=0 \\ x^2-6x+9 = (x-3)^2\\ y^2+2y+1 = (y+1)^2\\ z^2+8z+16 = (z+4)^2 ,then\\ (x-3)^2 - 9 +(y+1)^2 - 1+(z+4)^2-16-4 = 0\\ (x-3)^2 +(y+1)^2 +(z+4)^2 = 30\\ (x-a)^2 +(y-b)^2 +(z-c)^2 = R^2, then\\ a = 3,b=-1,c=-4, R^2 = 30$

Answer:

Equation of a sphere: $(x-3)^2 +(y+1)^2 +(z+4)^2 = 30\\$

Sphere center С(3, -1, -4)

Radius $\sqrt{30}$