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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)\\n

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

$$\mathrm{step}\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}\:? \\ $$ $$\color{mathred}{\left(}\mathrm{\color{mathred}{1}}\color{mathred}{+}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{2}}} \color{mathred}{+}\mathrm{\color{mathred}{3}}^{\mathrm{\color{mathred}{2}}} \color{mathred}{−}\mathrm{\color{mathred}{2}}\color{mathred}{\left(}\mathrm{\color{mathred}{1}}\color{mathred}{+}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{\right)}\color{mathred}{−}\mathrm{\color{mathred}{4}}\color{mathred}{.}\mathrm{\color{mathred}{3}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:} \\ $$ $$\color{mathred}{\:}\mathrm{\color{mathred}{6}}\color{mathred}{+}\mathrm{\color{mathred}{2}}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{\:}\color{mathred}{+}\color{mathred}{\:}\mathrm{\color{mathred}{9}}\color{mathred}{−}\mathrm{\color{mathred}{2}}\color{mathred}{−}\mathrm{\color{mathred}{2}}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{\:}\color{mathred}{−}\mathrm{\color{mathred}{1}\color{mathred}{2}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathblue}{1}}\color{mathblue}{\:}\color{mathblue}{\left(}\mathrm{\color{mathblue}{y}\color{mathblue}{e}\color{mathblue}{s}}\color{mathblue}{\:}\mathrm{\color{mathblue}{i}\color{mathblue}{t}}\color{mathblue}{\:}\mathrm{\color{mathblue}{i}\color{mathblue}{s}}\color{mathblue}{\:}\mathrm{\color{mathblue}{o}\color{mathblue}{n}}\color{mathblue}{\:}\mathrm{\color{mathblue}{t}\color{mathblue}{h}\color{mathblue}{e}}\color{mathblue}{\:}\mathrm{\color{mathblue}{c}\color{mathblue}{i}\color{mathblue}{r}\color{mathblue}{c}\color{mathblue}{l}\color{mathblue}{e}}\color{mathblue}{\:}\color{mathblue}{\right)} \\ $$ $$\mathrm{step}\left(\mathrm{2}\right)\:\mathrm{find}\:\mathrm{slope}\:\mathrm{the}\:\mathrm{line}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\: \\ $$ $$\mathrm{to}\:\mathrm{the}\:\mathrm{given}\:\mathrm{point}\:\Rightarrow\:\mathrm{m}=\frac{\mathrm{3}−\mathrm{2}}{\mathrm{1}+\sqrt{\mathrm{5}}−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$ $$\mathrm{step}\left(\mathrm{3}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\: \\ $$ $$\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{m}}\:=\:−\sqrt{\mathrm{5}} \\ $$ $$\mathrm{\color{mathred}{s}\color{mathred}{t}\color{mathred}{e}\color{mathred}{p}}\color{mathred}{\left(}\mathrm{\color{mathred}{4}}\color{mathred}{\right)}\color{mathred}{\:}\mathrm{\color{mathred}{w}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{g}\color{mathred}{e}\color{mathred}{t}}\color{mathred}{\:}\mathrm{\color{mathred}{t}\color{mathred}{h}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{t}\color{mathred}{a}\color{mathred}{n}\color{mathred}{g}\color{mathred}{e}\color{mathred}{n}\color{mathred}{t}}\color{mathred}{\:}\mathrm{\color{mathred}{h}\color{mathred}{a}\color{mathred}{s}}\color{mathred}{\:}\mathrm{\color{mathred}{e}\color{mathred}{q}\color{mathred}{u}\color{mathred}{a}\color{mathred}{t}\color{mathred}{i}\color{mathred}{o}\color{mathred}{n}} \\ $$ $$\mathrm{\color{mathred}{y}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\color{mathred}{−}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{\:}\color{mathred}{\left(}\mathrm{\color{mathred}{x}}\color{mathred}{−}\mathrm{\color{mathred}{1}}\color{mathred}{−}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{\right)}\color{mathred}{\:}\color{mathred}{+}\color{mathred}{\:}\mathrm{\color{mathred}{3}} \\ $$ $$\mathrm{\color{mathblue}{y}}\color{mathblue}{=}\color{mathblue}{−}\mathrm{\color{mathblue}{x}}\sqrt{\mathrm{\color{mathblue}{5}}}\color{mathblue}{\:}\color{mathblue}{+}\sqrt{\mathrm{\color{mathblue}{5}}}\color{mathblue}{\:}\color{mathblue}{+}\color{mathblue}{\:}\mathrm{\color{mathblue}{8}}\color{mathblue}{\:} \\ $$

Commented bybemath 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.

$$\boldsymbol{\mathrm{step}}\:\left(\mathrm{1}\right)\::\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{centre}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{cercle}}\:\left(\mathscr{C}\:\right)\:\boldsymbol{\mathrm{given}}. \\ $$ $$\left(\mathscr{C}\right)\::\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{4}\boldsymbol{{y}}\:=\:\mathrm{1}\:\Rightarrow\: \\ $$ $$\boldsymbol{\mathrm{I}}\left(\frac{\mathrm{2}}{\mathrm{2}}\:;\:\frac{\mathrm{4}}{\mathrm{2}}\right)\:\Rightarrow \\ $$ $$\boldsymbol{\mathrm{I}}\left(\mathrm{1}\:;\:\mathrm{2}\right). \\ $$ $$\boldsymbol{\mathrm{step}}\:\left(\mathrm{2}\right)\::\:\boldsymbol{\mathrm{M}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)\:\in\:\left(\mathscr{D}\:\right)\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{A}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\:;\:\mathrm{3}\right) \\ $$ $$\overset{\rightarrow} {\boldsymbol{\mathrm{IA}}}.\overset{\rightarrow} {\boldsymbol{\mathrm{AM}}}\:=\:\overset{\rightarrow} {\mathrm{0}}\:\Leftrightarrow \\ $$ $$\left(\mathrm{1}+\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{1}−\sqrt{\mathrm{5}}\right)+\left(\mathrm{3}−\mathrm{2}\right)\left(\boldsymbol{{y}}−\mathrm{3}\right)\:=\:\:\mathrm{0}\:\Leftrightarrow \\ $$ $$\sqrt{\mathrm{5}}\left(\boldsymbol{{x}}−\mathrm{1}−\sqrt{\mathrm{5}}\right)+\boldsymbol{{y}}−\mathrm{3}\:=\:\mathrm{0}\:\Leftrightarrow \\ $$ $$\boldsymbol{{x}}\sqrt{\mathrm{5}}+\boldsymbol{{y}}−\sqrt{\mathrm{5}}−\mathrm{5}−\mathrm{3}\:=\:\mathrm{0}\:\Leftrightarrow \\ $$ $$\boldsymbol{{x}}\sqrt{\mathrm{5}}+\boldsymbol{{y}}−\sqrt{\mathrm{5}}−\mathrm{8}\:=\:\mathrm{0}\:\Leftrightarrow \\ $$ $$\boldsymbol{{\color{mathred}{x}}}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{+}\boldsymbol{{\color{mathred}{y}}}\color{mathred}{−}\color{mathred}{\left(}\sqrt{\mathrm{\color{mathred}{5}}}\color{mathred}{+}\mathrm{\color{mathred}{8}}\color{mathred}{\right)}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{0}}. \\ $$

Commented bybemath last updated on 24/Mar/21

via vector

$${via}\:{vector} \\ $$

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