Find the equation of the sphere through the circle x^2+y^2=4;z=0and its intersected by the plane x+2y+2z=0is a of raduis 3
Answers
Answer:
general equation of sphere,
x
2
+y
2
+2gx+2fy+2hz+c=0
centre (−g,−f.−h) and radius =
g
2
+f
2
+h
2
−c
At z=0, x
2
+y
2
+2gx+2fy+c=0−−−(i)
x
2
+y
2
−4=0−−−(ii)
comparing (i) and (ii), g=0,f=0,c=−4
∴ general equation reduces to,
x
2
+y
2
+z
2
+2hz+4=0 centre (0,0,h), radius=
h
2
−c
from fig., P is perpendicular drawn from centre to plane and r is radius of sphere. ∴p
2
+9=r
2
9
4h
2
+9=h
2
+4
⇒h=±3
∴ equation of spheres are,
x
2
+y
2
+z
2
+6z−4=0 and x
2
+y
2
+z
2
+6z−4=0
Step-by-step explanation:
HOPE THIS IS HELPFUL
Answer:
The given circle is x
2
+y
2
−4=0,z=0
The equation of this circle can be written as x
2
+y
2
+z
2
−4=0, z=0
Any sphere through this circle is x
2
+y
2
+z
2
−4+kz=0 ...(1)
Its centre is C (0,0,
2
−k
) and radius is
4
k
2
+4
= CP
Now the sphere (1) is cut by the plane x+2y+2z=0 ...(2)
In a circle of radius 3, draw CA⊥ distance of (0,0,
2
−k
) from the plane (2) =
1+4+4
∣0+0−k∣
=
3
k
Now from the right ∠d△CAP,CA
2
+AP
2
=CP
2
⇒
9
k
2
+9=
4
k
2
+4
⇒k
2
=36
⇒k=±6
Putting these values of k in (1), the required sphere are x
2
+y
2
+z
2
±6z−4=0
Step-by-step explanation: