The equation of sphere which
passes through the origin and makes intercepts 3, 4 and 5 on the co-ordinate axes, is given by:
(a) x² + y² + z² - 3x – 4y + 5z = 0
(b) x² + y² + z² – 3x – 4y - 5z = 0
(c) x² + y² + z² - 4x - 5y - 3z = 0
(d) x² + y² + z² - 3x + 4y - 5z = 0
Answers
Answered by
0
Answer:
x2+y2+z2 -4x-5x-3z=0
Answered by
2
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:
please vote or thank you me
Similar questions