Equation of circle through 3 points (a,0,0), (0,b,0) and (0,0,c)
Answers
Circle in 3 dimensions :
In 3 dimensions, a circle is formed when a sphere is cut by a plane.
Sphere : x² + y² + z² + 2gx + 2fy + 2hz + d = 0
Plane : ax + by + cz + k = 0
Then, the equation of the circle containing the sphere and the plane can be determined as
x² + y² + z² + 2gx + 2fy + 2hz + d = 0 ,
ax + by + cz + k = 0
Solution :
Here, three points are given (a, 0, 0), (0, b, 0) and (0, 0, c).
Step 1 :
Finding the equation of the sphere
Let, the equation of the sphere be
x² + y² + z² + 2gx + 2fy + 2hz = 0 ...(1)
Since, the sphere (1) passes through the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c), so
d = 0 ...(i)
a² + 2ag + d = 0
or, a² + 2ag = 0 ...(ii)
b² + 2bf = 0 ...(iii) , since d = 0
c² + 2ch = 0 ...(iv) , since d = 0
Since a, b, c ≠ 0, from (ii), (iii) and (iv), we can find
g = -a/2 , f = -b/2 , h = -c/2
Thus, from (1), we find the sphere as
S : x² + y² + z² - ax - by - cz = 0
Step 2 :
Finding the equation of the plane
Let, the equation of the plane be
Ax + By + Cz + D = 0 ...(2)
Since, the plane (2) passes through the points (a, 0, 0), (0, b, 0), (0, 0, c), so
Aa + D = 0 ...(v)
Bb + D = 0 ...(vi)
Cc + D = 0 ...(vii)
From (v), (vi), (vii), we get
A = -D/a , B = -D/b , C = -D/c ,
where a, b, c ≠ 0
Thus, from (2), we find the required plane as
P : x/a + y/b + z/c = 1
Step 3 :
Finding the required equation of the circle
Hence, the required circle is represented by
S , P
i.e., x² + y² + z² - ax - by - cz = 0 ,
x/a + y/b + z/c = 1