Question Number 149796 by mr W last updated on 07/Aug/21

Commented bymr W last updated on 07/Aug/21

solution to Q148806

Commented bymr W last updated on 07/Aug/21

plane (x/a)+(y/b)+(z/c)=1 intersects the  coordinate axes at A,B,C.  A(a,0,0)  B(0,b,0)  A(0,0,c)  center of circumcircle of ΔABC is D.  D(α,β,γ)  we get from Q149428:  α=((a^3 (b^2 +c^2 ))/(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )))  β=((b^3 (c^2 +a^2 ))/(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )))  γ=((c^3 (a^2 +b^2 ))/(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )))  radius of circle is r.  r=(1/2)(√(((a^2 +b^2 )(b^2 +c^2 )(c^2 +a^2 ))/(a^2 b^2 +b^2 c^2 +c^2 a^2 )))    OD^(→) =(α,β,γ)  normal of plane n^(→) =((1/a),(1/b),(1/c))  say u^(→) ,v^(→)  are two perpendicular unit   vectors in the plane.  OP^(→) =OD^(→) +DP^(→) =OD^(→) +r cos θ u^(→) +r sin θ v^(→)   u^→  is ⊥ n^(→) ,  u_x ×(1/a)+u_y ×(1/b)+u_z ×(1/c)=0  for example u_x =a, u_y =−b, u_z =0  i.e. u^(→)  // AB^(→)   v^→ =n^(→) ×u^(→)   = determinant (((1/a),((1/b) ),(1/c)),(a,(−b),0))=((b/c),(a/c),−(a/b)−(b/a))  since u^(→) , v^(→)  are unit vectors,  u^→ =((a/( (√(a^2 +b^2 )))),−(b/( (√(a^2 +b^2 )))),0)  v^→ =((b/( c(√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))),(a/( c(√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))),−(((a^2 +b^2 ))/( ab(√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))))  OP^(→) =(α,β,γ)+(r/( (√(a^2 +b^2 ))))(a,−b, 0) cos θ+(r/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) ((b/c),(a/c),−((a^2 +b^2 )/(ab)))sin θ  or   { ((x=α+((ar)/( (√(a^2 +b^2 ))))cos θ+((((b/c))r)/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) sin θ)),((y=β−((br)/( (√(a^2 +b^2 ))))cos θ+((((a/c))r)/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) sin θ)),((z=γ−((((a/b)+(b/a))r)/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) sin θ)) :}  this is the equation of circumcircle  with θ as parameter.

Commented byTawa11 last updated on 07/Aug/21

Kudox sir

Commented bymr W last updated on 07/Aug/21