Given the plane:
x−2y+2z−6=0
and the sphere equation:
x2+y2+z2−3x−6z+9=0
⇒(x−23)2+(y−0)2+(z−3)2=(23)2by completing the square method
Then the center of the sphere is (23, 0, 3) with radius 23.
Now, the length of the perpendicular from (23, 0, 3) to the plane x−2y+2z−6=0 is;
OM=a2+b2+c2∣ax1+by1+cz1+d∣
Here, a=1,b=−2,c=2,d=−6,x1=23,y1=0,z1=3. Thus,
OM=12+(−2)2+22∣1⋅23+(−2)⋅0+2⋅3+(−6)∣=3∣23∣=21
Next, AM2=OA2−OM2=(23)2−(21)2=2
The equation of the axis of the cylinder is;
lx−α=my−β=nx−γ, Here, (α, β, γ)=(23, 0, 3), l=1, m=−2, n=2.
⇒1x−23=−2y−0=2x−3, and l2+m2+n2=9
The equation of the Right Circular Cylinder is;
[n(y−β)−m(z−γ)]2+[l(z−γ)−n(x−α)]2+[m(x−α)−l(y−β)]2=AM2(l2+m2+n2)⇒[2(y−0)+2(z−3)]2+[1(z−3)−2(x−23)]2+[−2(x−23)−1(y−0)]2=2(9)⇒[2y+2z−6)]2+[z−2x]2+[−2x+3−y]2=18⇒8x2+5y2+5z2+8yz−4xz+4xy−12x−30y−24z+27=0
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