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Question Number 136604 by bemath last updated on 23/Mar/21
What is the equation of a tangent to a circle whose equation is x^2+y^2−2x−4y=1 at point (1+√5,3)
$$ \\ $$What is the equation of a tangent to a circle whose equation is x^2+y^2−2x−4y=1 at point (1+√5,3)
Answered by EDWIN88 last updated on 23/Mar/21
step(1) is the point on the circle ? (1+(√5))^2 +3^2 −2(1+(√5))−4.3 = 6+2(√5) + 9−2−2(√5) −12 = 1 (yes it is on the circle ) step(2) find slope the line from the center to the given point ⇒ m=((3−2)/(1+(√5)−1)) = (1/( (√5))) step(3) find the slope of tangent line is −(1/m) = −(√5) step(4) we get the tangent has equation y = −(√5) (x−1−(√5)) + 3 y=−x(√5) +(√5) + 8
$$\mathrm{step}\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}\:? \\ $$$$\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{4}.\mathrm{3}\:=\: \\ $$$$\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:+\:\mathrm{9}−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{12}\:=\:\mathrm{1}\:\left(\mathrm{yes}\:\mathrm{it}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}\:\right) \\ $$$$\mathrm{step}\left(\mathrm{2}\right)\:\mathrm{find}\:\mathrm{slope}\:\mathrm{the}\:\mathrm{line}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\: \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{given}\:\mathrm{point}\:\Rightarrow\:\mathrm{m}=\frac{\mathrm{3}−\mathrm{2}}{\mathrm{1}+\sqrt{\mathrm{5}}−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{step}\left(\mathrm{3}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\: \\ $$$$\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{m}}\:=\:−\sqrt{\mathrm{5}} \\ $$$$\mathrm{step}\left(\mathrm{4}\right)\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{has}\:\mathrm{equation} \\ $$$$\mathrm{y}\:=\:−\sqrt{\mathrm{5}}\:\left(\mathrm{x}−\mathrm{1}−\sqrt{\mathrm{5}}\right)\:+\:\mathrm{3} \\ $$$$\mathrm{y}=−\mathrm{x}\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{5}}\:+\:\mathrm{8}\: \\ $$
Commented by bemath last updated on 24/Mar/21
Answered by greg_ed last updated on 24/Mar/21
step (1) : find the centre I of the cercle (C ) given. (C) : x^2 +y^2 −2x−4y = 1 ⇒ I((2/2) ; (4/2)) ⇒ I(1 ; 2). step (2) : M(x;y) ∈ (D ) the tangent and A(1+(√5) ; 3) IA^(→) .AM^(→) = 0^→ ⇔ (1+(√5)−1)(x−1−(√5))+(3−2)(y−3) = 0 ⇔ (√5)(x−1−(√5))+y−3 = 0 ⇔ x(√5)+y−(√5)−5−3 = 0 ⇔ x(√5)+y−(√5)−8 = 0 ⇔ x(√5)+y−((√5)+8) = 0.
$$\boldsymbol{\mathrm{step}}\:\left(\mathrm{1}\right)\::\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{centre}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{cercle}}\:\left(\mathscr{C}\:\right)\:\boldsymbol{\mathrm{given}}. \\ $$$$\left(\mathscr{C}\right)\::\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{4}\boldsymbol{{y}}\:=\:\mathrm{1}\:\Rightarrow\: \\ $$$$\boldsymbol{\mathrm{I}}\left(\frac{\mathrm{2}}{\mathrm{2}}\:;\:\frac{\mathrm{4}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\boldsymbol{\mathrm{I}}\left(\mathrm{1}\:;\:\mathrm{2}\right). \\ $$$$\boldsymbol{\mathrm{step}}\:\left(\mathrm{2}\right)\::\:\boldsymbol{\mathrm{M}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)\:\in\:\left(\mathscr{D}\:\right)\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{A}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\:;\:\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{IA}}}.\overset{\rightarrow} {\boldsymbol{\mathrm{AM}}}\:=\:\overset{\rightarrow} {\mathrm{0}}\:\Leftrightarrow \\ $$$$\left(\mathrm{1}+\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{1}−\sqrt{\mathrm{5}}\right)+\left(\mathrm{3}−\mathrm{2}\right)\left(\boldsymbol{{y}}−\mathrm{3}\right)\:=\:\:\mathrm{0}\:\Leftrightarrow \\ $$$$\sqrt{\mathrm{5}}\left(\boldsymbol{{x}}−\mathrm{1}−\sqrt{\mathrm{5}}\right)+\boldsymbol{{y}}−\mathrm{3}\:=\:\mathrm{0}\:\Leftrightarrow \\ $$$$\boldsymbol{{x}}\sqrt{\mathrm{5}}+\boldsymbol{{y}}−\sqrt{\mathrm{5}}−\mathrm{5}−\mathrm{3}\:=\:\mathrm{0}\:\Leftrightarrow \\ $$$$\boldsymbol{{x}}\sqrt{\mathrm{5}}+\boldsymbol{{y}}−\sqrt{\mathrm{5}}−\mathrm{8}\:=\:\mathrm{0}\:\Leftrightarrow \\ $$$$\boldsymbol{{x}}\sqrt{\mathrm{5}}+\boldsymbol{{y}}−\left(\sqrt{\mathrm{5}}+\mathrm{8}\right)\:=\:\mathrm{0}. \\ $$
Commented by bemath last updated on 24/Mar/21
via vector
$${via}\:{vector} \\ $$

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