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

Commented by mr W last updated on 07/Aug/21

$s o l u t i o n t o Q 148806$
Commented by mr W last updated on 07/Aug/21
$plane (x/a)+(y/b)+(z/c)=1 intersects the coordinate axes at A,B,C. A(a,0,0) B(0,b,0) A(0,0,c) center of circumcircle of ΔABC is D. D(α,β,γ) we get from Q149428: α=((a^3 (b^2 +c^2 ))/(2(a^2 b^2 +b^2 c^2 +c^2 a^2 ))) β=((b^3 (c^2 +a^2 ))/(2(a^2 b^2 +b^2 c^2 +c^2 a^2 ))) γ=((c^3 (a^2 +b^2 ))/(2(a^2 b^2 +b^2 c^2 +c^2 a^2 ))) radius of circle is r. r=(1/2)(√(((a^2 +b^2 )(b^2 +c^2 )(c^2 +a^2 ))/(a^2 b^2 +b^2 c^2 +c^2 a^2 ))) OD^(→) =(α,β,γ) normal of plane n^(→) =((1/a),(1/b),(1/c)) say u^(→) ,v^(→) are two perpendicular unit vectors in the plane. OP^(→) =OD^(→) +DP^(→) =OD^(→) +r cos θ u^(→) +r sin θ v^(→) u^→ is ⊥ n^(→) , u_x ×(1/a)+u_y ×(1/b)+u_z ×(1/c)=0 for example u_x =a, u_y =−b, u_z =0 i.e. u^(→) // AB^(→) v^→ =n^(→) ×u^(→) = determinant (((1/a),((1/b) ),(1/c)),(a,(−b),0))=((b/c),(a/c),−(a/b)−(b/a)) since u^(→) , v^(→) are unit vectors, u^→ =((a/( (√(a^2 +b^2 )))),−(b/( (√(a^2 +b^2 )))),0) v^→ =((b/( c(√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))),(a/( c(√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))),−(((a^2 +b^2 ))/( ab(√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 ))))))) OP^(→) =(α,β,γ)+(r/( (√(a^2 +b^2 ))))(a,−b, 0) cos θ+(r/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) ((b/c),(a/c),−((a^2 +b^2 )/(ab)))sin θ or { ((x=α+((ar)/( (√(a^2 +b^2 ))))cos θ+((((b/c))r)/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) sin θ)),((y=β−((br)/( (√(a^2 +b^2 ))))cos θ+((((a/c))r)/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) sin θ)),((z=γ−((((a/b)+(b/a))r)/( (√((a^2 +b^2 )((1/a^2 )+(1/b^2 )+(1/c^2 )))))) sin θ)) :} this is the equation of circumcircle with θ as parameter.$
$p l a n e \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 i n t e r s e c t s t h e$ $c o o r d i n a t e a x e s a t A, B, C .$ $A (a, 0, 0)$ $B (0, b, 0)$ $A (0, 0, c)$ $c e n t e r o f c i r c u m c i r c l e o f Δ A B C i s D .$ $D (α, β, γ)$ $w e g e t f r o m Q 149428 :$ $α = \frac{a^{3} (b^{2} + c^{2})}{2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})}$ $β = \frac{b^{3} (c^{2} + a^{2})}{2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})}$ $γ = \frac{c^{3} (a^{2} + b^{2})}{2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})}$ $r a d i u s o f c i r c l e i s r .$ $r = \frac{1}{2} \sqrt{\frac{(a^{2} + b^{2}) (b^{2} + c^{2}) (c^{2} + a^{2})}{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}}}$ $\vec{O D} = (α, β, γ)$ $n o r m a l o f p l a n e \vec{n} = (\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$ $s a y \vec{u}, \vec{v} a r e t w o p e r p e n d i c u l a r u n i t$ $v e c t o r s i n t h e p l a n e .$ $\vec{O P} = \vec{O D} + \vec{D P} = \vec{O D} + r \cos θ \vec{u} + r \sin θ \vec{v}$ $\vec{u} i s ⊥ \vec{n},$ $u_{x} \times \frac{1}{a} + u_{y} \times \frac{1}{b} + u_{z} \times \frac{1}{c} = 0$ $f o r e x a m p l e u_{x} = a, u_{y} = - b, u_{z} = 0$ $i . e . \vec{u} / / \vec{A B}$ $\vec{v} = \vec{n} \times \vec{u}$ $= \| \begin{matrix} \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ a & - b & 0 \end{matrix} \| = (\frac{b}{c}, \frac{a}{c}, - \frac{a}{b} - \frac{b}{a})$ $s i n c e \vec{u}, \vec{v} a r e u n i t v e c t o r s,$ $\vec{u} = (\frac{a}{\sqrt{a^{2} + b^{2}}}, - \frac{b}{\sqrt{a^{2} + b^{2}}}, 0)$ $\vec{v} = (\frac{b}{c \sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}}, \frac{a}{c \sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}}, - \frac{(a^{2} + b^{2})}{a b \sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}})$ $\vec{O P} = (α, β, γ) + \frac{r}{\sqrt{a^{2} + b^{2}}} (a, - b, 0) \cos θ + \frac{r}{\sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}} (\frac{b}{c}, \frac{a}{c}, - \frac{a^{2} + b^{2}}{a b}) \sin θ$ $o r$ ${\begin{cases} x = α + \frac{a r}{\sqrt{a^{2} + b^{2}}} \cos θ + \frac{(\frac{b}{c}) r}{\sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}} \sin θ \\ y = β - \frac{b r}{\sqrt{a^{2} + b^{2}}} \cos θ + \frac{(\frac{a}{c}) r}{\sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}} \sin θ \\ z = γ - \frac{(\frac{a}{b} + \frac{b}{a}) r}{\sqrt{(a^{2} + b^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}} \sin θ \end{cases}$ $t h i s i s t h e e q u a t i o n o f c i r c u m c i r c l e$ $w i t h θ a s p a r a m e t e r .$
Commented by Tawa11 last updated on 07/Aug/21

$Kudox sir$
Commented by mr W last updated on 07/Aug/21

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