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(1 point) Answer the following questions about the sphere whose equation is given by x2+y2+z2−4x+4y=73. x2+y2+z2−4x+4y=73. 1. Find the radius of the sphere. Radius: r=r= 2. Find the center of the sphere. Write the center as a point (a,b,c)(a,b,c) where aa, bb, and cc are numbers.

Respuesta :

AMB000

Answer:

R=9.

C=(2,-2,0).

Explanation:

Lets try to find how [tex]x^2+y^2+z^2-4x+4y=73[/tex] can be written in the form of [tex](x-a)^2+(x-b)^2+(x-c)^2=R^2[/tex], since then its radius R and center of coordinates C=(a,b,c) would be easily recognizable.

Since we have the terms -4x and +4y, they seem to have come from the terms [tex](x-2)^2[/tex] and [tex](y+2)^2[/tex], and we would need to take care of the extra +4 by substracting then. More explicitly:

[tex](x-2)^2-4=x^2-4x+4-4=x^2-4x[/tex], as we had originally with the x terms

[tex](y+2)^2-4=y^2+4y+4-4=y^2+4y[/tex], as we had originally with the y terms

Which means:

[tex]x^2-4x+y^2+4y+z^2=(x-2)^2-4+(y+2)^2-4+z^2[/tex]

Or, putting all together:

[tex](x-2)^2-4+(y+2)^2-4+z^2=73[/tex]

[tex](x-2)^2+(y+2)^2+z^2=9^2[/tex]

So our radius is 9 and the center is (2,-2,0).

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