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Question Number 135000 by bobhans last updated on 09/Mar/21

$$ \\ $$ Find the equation of the circle through the points of intersection of x^2+y^2−1=0,x^2+y^2−2x−4y+1=0 and touching the line x+2y=0?\\n

Answered by EDWIN88 last updated on 09/Mar/21

Let C is equation of circle , so C≡ x^2 +y^2 −1+λ(−2x−4y+2)=0 ≡ x^2 +y^2 −2λx−4λy+2λ−1 = 0 witb center point at (λ,2λ) and radius r = (√(λ^2 +4λ^2 +1−2λ)) = ((∣λ+4λ∣)/( (√5))) ⇒(√(5(5λ^2 +1−2λ))) = ∣5λ∣ ⇒25λ^2 +5−10λ = 25λ^2 ; λ=(1/2) so we get solution ≡ x^2 +y^2 −x−2y = 0

$$\mathrm{Let}\:\mathrm{C}\:\mathrm{is}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\:,\:\mathrm{so}\: \\ $$ $$\mathrm{C}\equiv\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{1}+\lambda\left(−\mathrm{2x}−\mathrm{4y}+\mathrm{2}\right)=\mathrm{0} \\ $$ $$\equiv\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2}\lambda\mathrm{x}−\mathrm{4}\lambda\mathrm{y}+\mathrm{2}\lambda−\mathrm{1}\:=\:\mathrm{0} \\ $$ $$\mathrm{witb}\:\mathrm{center}\:\mathrm{point}\:\mathrm{at}\:\left(\lambda,\mathrm{2}\lambda\right)\:\mathrm{and}\:\mathrm{radius}\: \\ $$ $$\mathrm{r}\:=\:\sqrt{\lambda^{\mathrm{2}} +\mathrm{4}\lambda^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\lambda}\:=\:\frac{\mid\lambda+\mathrm{4}\lambda\mid}{\:\sqrt{\mathrm{5}}} \\ $$ $$\Rightarrow\sqrt{\mathrm{5}\left(\mathrm{5}\lambda^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\lambda\right)}\:=\:\mid\mathrm{5}\lambda\mid \\ $$ $$\Rightarrow\mathrm{25}\lambda^{\mathrm{2}} +\mathrm{5}−\mathrm{10}\lambda\:=\:\mathrm{25}\lambda^{\mathrm{2}} \:;\:\lambda=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{\color{mathred}{s}\color{mathred}{o}}\color{mathred}{\:}\mathrm{\color{mathred}{w}\color{mathred}{e}}\color{mathred}{\:}\mathrm{\color{mathred}{g}\color{mathred}{e}\color{mathred}{t}}\color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{o}\color{mathred}{l}\color{mathred}{u}\color{mathred}{t}\color{mathred}{i}\color{mathred}{o}\color{mathred}{n}}\color{mathred}{\:}\color{mathred}{\equiv}\color{mathred}{\:}\mathrm{\color{mathred}{x}}^{\mathrm{\color{mathred}{2}}} \color{mathred}{+}\mathrm{\color{mathred}{y}}^{\mathrm{\color{mathred}{2}}} \color{mathred}{−}\mathrm{\color{mathred}{x}}\color{mathred}{−}\mathrm{\color{mathred}{2}\color{mathred}{y}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{0}} \\ $$

Commented bybobhans last updated on 09/Mar/21

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