Question

Equation of circle through 3 points (a,0,0), (0,b,0) and (0,0,c)

Answer 1

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