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Find-the-equation-and-radius-of-the-circumcircle-of-the-triangle-formed-by-the-three-line-2y-9x-26-0-9y-2x-32-0-11y-7x-27-0-

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Question Number 15868 by tawa tawa last updated on 14/Jun/17
Find the equation and radius of the circumcircle of the triangle formed by the three line: 2y − 9x + 26 = 0 9y + 2x + 32 = 0 11y − 7x − 27 = 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{formed}\:\mathrm{by}\: \\ $$$$\mathrm{the}\:\mathrm{three}\:\mathrm{line}: \\ $$$$\mathrm{2y}\:−\:\mathrm{9x}\:+\:\mathrm{26}\:=\:\mathrm{0} \\ $$$$\mathrm{9y}\:+\:\mathrm{2x}\:+\:\mathrm{32}\:=\:\mathrm{0} \\ $$$$\mathrm{11y}\:−\:\mathrm{7x}\:−\:\mathrm{27}\:=\:\mathrm{0} \\ $$
Answered by Tinkutara last updated on 14/Jun/17
2y − 9x + 26 = 0 ...(i) 9y + 2x + 32 = 0 ...(ii) 11y − 7x − 27 = 0 ...(iii) Solving (i) and (ii), A = (2, −4) Solving (ii) and (iii), C = (−7, −2) Solving (iii) and (i), B = (4, 5) Slope of (i) = (9/2) and that of (ii) = ((−2)/9) so AB ⊥ AC and ΔABC is right angled at A. ∴ Circumcenter = midpoint of BC and circumradius = ((BC)/2) . Circumcenter = (−1.5, 1.5) and circumradius = ((√(170))/2) Equation of circumcircle = (x + 1.5)^2 + (y − 1.5)^2 = ((170)/4) = 42.5
$$\mathrm{2}{y}\:−\:\mathrm{9}{x}\:+\:\mathrm{26}\:=\:\mathrm{0}\:…\left({i}\right) \\ $$$$\mathrm{9}{y}\:+\:\mathrm{2}{x}\:+\:\mathrm{32}\:=\:\mathrm{0}\:…\left({ii}\right) \\ $$$$\mathrm{11}{y}\:−\:\mathrm{7}{x}\:−\:\mathrm{27}\:=\:\mathrm{0}\:…\left({iii}\right) \\ $$$$\mathrm{Solving}\:\left({i}\right)\:\mathrm{and}\:\left({ii}\right),\:{A}\:=\:\left(\mathrm{2},\:−\mathrm{4}\right) \\ $$$$\mathrm{Solving}\:\left({ii}\right)\:\mathrm{and}\:\left({iii}\right),\:{C}\:=\:\left(−\mathrm{7},\:−\mathrm{2}\right) \\ $$$$\mathrm{Solving}\:\left({iii}\right)\:\mathrm{and}\:\left({i}\right),\:{B}\:=\:\left(\mathrm{4},\:\mathrm{5}\right) \\ $$$$\mathrm{Slope}\:\mathrm{of}\:\left({i}\right)\:=\:\frac{\mathrm{9}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:\left({ii}\right)\:=\:\frac{−\mathrm{2}}{\mathrm{9}} \\ $$$$\mathrm{so}\:{AB}\:\bot\:{AC}\:\mathrm{and}\:\Delta{ABC}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled} \\ $$$$\mathrm{at}\:{A}.\:\therefore\:\mathrm{Circumcenter}\:=\:\mathrm{midpoint}\:\mathrm{of}\:{BC} \\ $$$$\mathrm{and}\:\mathrm{circumradius}\:=\:\frac{{BC}}{\mathrm{2}}\:. \\ $$$$\mathrm{Circumcenter}\:=\:\left(−\mathrm{1}.\mathrm{5},\:\mathrm{1}.\mathrm{5}\right)\:\mathrm{and} \\ $$$$\mathrm{circumradius}\:=\:\frac{\sqrt{\mathrm{170}}}{\mathrm{2}} \\ $$$$\mathrm{Equation}\:\mathrm{of}\:\mathrm{circumcircle}\:= \\ $$$$\left({x}\:+\:\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} \:+\:\left({y}\:−\:\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{170}}{\mathrm{4}}\:=\:\mathrm{42}.\mathrm{5} \\ $$
Commented by tawa tawa last updated on 14/Jun/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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