Answer to Question #91543 in Analytic Geometry for Ra

Question

Find the equations of the spheres which pass through the circle x²+y²+z²=9, 2x+2y-7=0 and touch the plane x-y+z+3=0

Answer 1

1. x²+y²+z²=9 is an equation of the sphere S₁ with center O₁(0,0,0) and radius$R_1=\sqrt{9}=3.$

2. The sign distance from the plane ax+by+cz+d=0 to the point X₀(x₀,y₀,z₀) is determined by

$$D=\frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}},$$

which is positive if X₀ is on the same side of the plane as the normal vector n=(a,b,c) and negative if it is on the opposite side. Hens, the distance from the plane 2x+2y-7=0 to the point O₁(0,0,0) is equal to

$$D_1=\frac{-7}{\sqrt{2^2+2^2+0^2}}=-\frac{7}{\sqrt{8}}.$$

3. The radius of the given circle can be determined by the Pythagorean theorem:

$$r_1=\sqrt{R_1^2-D_1^2}=\sqrt{3^2-(\frac{-7}{\sqrt{8}})^2}=\sqrt{9-\frac{49}{8}}=\sqrt{\frac{23}{8}}.$$

4. The sphere S₂ passes through the given circle, so its center O₂ lies on the line parallel to n₁ and passing through O₁:

$$O_2 = O_1 + λ\bold{n_1}=(0,0,0)+λ(2,2,0)=(2λ,2λ,0),$$

and its radius R₂ is equal to

$$D_{21}=\frac{2*2λ+2*2λ+0*0-7}{\sqrt{2^2+2^2+0^2}}=\frac{8λ-7}{\sqrt{8}};$$ $$R_2=\sqrt{r_1^2+D_{21}^2}=\sqrt{\frac{23}{8}+\frac{(8λ-7)^2}{8}}=\sqrt{\frac{23+(8λ-7)^2}{8}}.$$

5. Because the sphere S₂ toches the plane x-y+z+3=0, then R₂ is equal to the distance from its center to the plane:

$$D_2=\frac{1*2λ-1*2λ+1*0+3}{\sqrt{1^2+(-1)^2+1^2}}=\sqrt{3}=R_2;$$

$$\sqrt{\frac{23+(8λ-7)^2}{8}}=\sqrt{3};$$

$$23+(8λ-7)^2=24;$$

$$(8λ-7)^2=1;$$

$$8λ-7=±1;$$

$$8λ=7±1.$$

6. Equations of the spheres:

1) $λ_1=\frac{6}{8}=\frac{3}{4};$ $O_2^1 =(2λ_1,2λ_1,0)=(\frac{3}{2},\frac{3}{2},0);$

$(x-\frac{3}{2})²+(y-\frac{3}{2})²+z²=R_2^2=3;$

2) $λ_2=\frac{8}{8}=1;O_2^2 =(2λ_2,2λ_2,0)=(2,2,0);$

$(x-2)²+(y-2)²+z²=R_2^2=3.$

Answer:

$(x-\frac{3}{2})²+(y-\frac{3}{2})²+z²=3;$

$(x-2)²+(y-2)²+z²=3.$