Find out the centre and redius of circle in which the sphere x^2+y^2+z^2-8x+4y+8z-45=0 cut by the plane x-2y+2z=3
Answers
Answer:
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Step-by-step explanation:
Equation of sphere: x^2+y^2+z^2-8x+4y+8z-45=0x
2
+y
2
+z
2
−8x+4y+8z−45=0
Equation of plane: x-2y+2z=3x−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(x−4)
2
+(y+2)
2
+(z+4)
2
=81
Center: (4,-2,-4)
Normal vector to plane: x-2y+2z=3x−2y+2z=3
\vec{b}=<1,-2,2>
b
=<1,−2,2>
\vec{a}=<4,-2,-4>
a
=<4,−2,−4>
Equation of line:-
\vec{r}=<4,-2,-4>+t<1,-2,2>
r
=<4,−2,−4>+t<1,−2,2>
=<4+t,-2-2t,-4+2t>
Intersection of line and plane:-
4+t-2(-2-2t)+2(-4+2t)=34+t−2(−2−2t)+2(−4+2t)=3
4+t+4+4t-8+4t=34+t+4+4t−8+4t=3
9t=39t=3
t=\dfrac{1}{3}t=
3
1
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})(4+
3
1
,−2−
3
2
,−4+
3
2
)
Center of circle: \left(\frac{13}{3},-\frac{8}{3},-\frac{10}{3}\right)(
3
13
,−
3
8
,−
3
10
)
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=
(4−
3
13
)
2
+(−2+
3
8
)
2
+(−4+
3
10
)
2
d=1d=1
Radius of sphere: R = 9
Radius of circle: r=\sqrt{R^2-d^2}r=
R
2
−d
2
r=\sqrt{81-1}r=
81−1
r=\sqrt{80}r=
80
r=4\sqrt{5}r=4
5