Taliyah Reyes
2022-08-03
Answered

Find an equation for each sphere that passes through the point(5,1,4) and is tangent to all three coordinate planes.

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kunstboom8w

Answered 2022-08-04
Author has **8** answers

the general sphere equation is

$(x-{x}_{0}{)}^{2}+(y-{y}_{0}{)}^{2}+(z-{z}_{0}{)}^{2}={R}^{2}$...(1)

$({x}_{0},{y}_{0},{z}_{0})$ is the centerof sphere

since the sphere tangent to all three coordinate planes, we have:

${x}_{0}={y}_{0}={z}_{0}=R$

(1) change to:

$(x-R{)}^{2}+(y-R{)}^{2}+(z-R{)}^{2}=R$

sphere passes through the point (5,1,4) , we have:

$(5-R{)}^{2}+(1-R{)}^{2}+(4-R{)}^{2}=R$

$\Rightarrow 25-10R+{R}^{2}+1-2R+{R}^{2}+16-8R+{R}^{2}={R}^{2}$

$2{R}^{2}-20R+32=0$

${R}^{2}-10R+16=0$

(R-2)(R-8) = 0

R=2, or R=8

the sphere equations are

$(x-2{)}^{2}+(y-2{)}^{2}+(z-2{)}^{2}={2}^{2}$

or

$(x-8{)}^{2}+(y-8{)}^{2}+(z-8{)}^{2}={8}^{2}$

$(x-{x}_{0}{)}^{2}+(y-{y}_{0}{)}^{2}+(z-{z}_{0}{)}^{2}={R}^{2}$...(1)

$({x}_{0},{y}_{0},{z}_{0})$ is the centerof sphere

since the sphere tangent to all three coordinate planes, we have:

${x}_{0}={y}_{0}={z}_{0}=R$

(1) change to:

$(x-R{)}^{2}+(y-R{)}^{2}+(z-R{)}^{2}=R$

sphere passes through the point (5,1,4) , we have:

$(5-R{)}^{2}+(1-R{)}^{2}+(4-R{)}^{2}=R$

$\Rightarrow 25-10R+{R}^{2}+1-2R+{R}^{2}+16-8R+{R}^{2}={R}^{2}$

$2{R}^{2}-20R+32=0$

${R}^{2}-10R+16=0$

(R-2)(R-8) = 0

R=2, or R=8

the sphere equations are

$(x-2{)}^{2}+(y-2{)}^{2}+(z-2{)}^{2}={2}^{2}$

or

$(x-8{)}^{2}+(y-8{)}^{2}+(z-8{)}^{2}={8}^{2}$

popljuvao69

Answered 2022-08-05
Author has **2** answers

everything answered above is totally perfect except that 25+1+16=42, and not 25+1+16=32. So, radius equals 3 and 7.

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