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write the equation of the circle of the triangle with vertices $$A = (5 ,\ -4 ),\ B = (6 ,\ -1 ),\ C = ( 2,\ 3)$$ examine the relative position of this district and its image in axial symmetry about the line $$3x + 4y + 26 = 0$$

I think about this: (drawing paper)enter image description here

Jon.Don
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1 Answers1

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Substitute $(A,B,C)$ into $(x-a)^2 + (y-b)^2 = R^2$ : $$ (5-a)^2 + (-4-b)^2 = R^2 \\ (6-a)^2 + (-1-b)^2 = R^2 \\ (2-a)^2 + (3-b)^2 = R^2 $$ $$ a^2 - 10 a + 25 + b^2 + 8 b + 16 = R^2 \\ a^2 - 12 a + 36 + b^2 + 2 b + 1 = R^2 \\ a^2 - 4 a + 4 + b^2 - 6 b + 9 = R^2 $$ Subtract the second equation from the first one and the third equation from the second one: $$ 2 a - 11 + 6 b + 15 = 0 \\ - 8 a + 32 + 6 b - 3 = 0 $$ $$ 2 a + 6 b = - 4 \\ - 8 a + 8 b = - 24 $$ Add the second equation to $4 \times$ the first one .. and do the rest: $$ 32 b = - 40 \quad \Longrightarrow \quad b = - 5/4 \\ 2\cdot 2 a = - 8 + 15 = 7 \quad \Longrightarrow \quad a = 7/4 \\ R^2 = (2-a)^2 + (3-b)^2 = 145/8 $$ So the circle is: $$ (x-7/4)^2 + (y+5/4)^2 = \left(\sqrt{145/8}\right)^2 $$ enter image description here
Still don't know what you mean by "district" ..

Han de Bruijn
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  • Investigate the relative position of this circle and its image in the axial symmetry about the line 3x + 4y + 26 = 0 – Jon.Don Sep 13 '15 at 16:12
  • Onderzoeken de relatieve positie van de wijk en haar imago in axiale symmetrie over de lijn 3x + 4y + 26 = – Jon.Don Sep 13 '15 at 19:45
  • Google translate my friend :) – Jon.Don Sep 13 '15 at 19:45
  • @Jon.Don: Sure, so the translation is dreadful, which makes it even less comprehensible. English is not my problem, though. Manjana. – Han de Bruijn Sep 13 '15 at 19:57
  • hm I have explored the relative position of this circle and its reflections in the axial symmetry about the line 3x+4y+26 = 0 – Jon.Don Sep 13 '15 at 20:00