Question

Find the radius and centre of the circle. x2 y2 z2 – 8x 4y 8z – 45 = 0, x – 2y 2z = 3

Answer 1

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Step-by-step explanation:

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Answer 2

The radius of circle is $4\sqrt{5}$ unit

The center of circle is $\left(\frac{13}{3},-\frac{8}{3},-\frac{10}{3}\right)$

Step-by-step explanation:

• Equation of sphere: $x^2+y^2+z^2-8x+4y+8z-45=0$

• Equation of plane: $x-2y+2z=3$

The intersection of sphere and plane makes a circle.

The normal of the plane passing through center of sphere.

The equation of line perpendicular to plane and passing through center of sphere.

Center of sphere: $(x-4)^2+(y+2)^2+(z+4)^2=81$

Center: (4,-2,-4)

Normal vector to plane: $x-2y+2z=3$

$\vec{b}=<1,-2,2>$

$\vec{a}=<4,-2,-4>$

Equation of line:-

$\vec{r}=<4,-2,-4>+t<1,-2,2>$

$<x,y,z>=<4+t,-2-2t,-4+2t>$

Intersection of line and plane:-

$4+t-2(-2-2t)+2(-4+2t)=3$

$4+t+4+4t-8+4t=3$

$9t=3$

$t=\dfrac{1}{3}$

Point of intersection of plane and line is center of intersecting circle.

Center of circle: $(4+\dfrac{1}{3},-2-\dfrac{2}{3},-4+\dfrac{2}{3})$

Center of circle: $\left(\frac{13}{3},-\frac{8}{3},-\frac{10}{3}\right)$

Distance of plane from center of sphere:-

$d=\sqrt{\left(4-\frac{13}{3}\right)^{2}+\left(-2+\frac{8}{3}\right)^{2}+\left(-4+\frac{10}{3}\right)^{2}}$

$d=1$

Radius of sphere: R = 9

Radius of circle: $r=\sqrt{R^2-d^2}$

$r=\sqrt{81-1}$

$r=\sqrt{80}$

$r=4\sqrt{5}$

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