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Given-the-equations-of-twe-circles-C-1-x-2-y-2-6x-4y-9-0-and-C-2-x-2-y-2-2x-6y-9-a-Find-the-equation-of-the-circle-C-3-which-passes-through-the-centre-of-C-1-and-through-the

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Question Number 109075 by Rio Michael last updated on 20/Aug/20
Given the equations of twe circles C_1 : x^2 + y^2 −6x−4y + 9 = 0 and C_2 : x^2 + y^2 −2x−6y + 9. (a) Find the equation of the circle C_3 which passes through the centre of C_1 and through the point of intersection of C_1 and C_2 . (b) The equations of two tangents from the origin to C_1 and the lenght of each tangent.
$$\mathrm{Given}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{twe}\:\mathrm{circles}\: \\ $$$${C}_{\mathrm{1}} \::\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{6}{x}−\mathrm{4}{y}\:+\:\mathrm{9}\:=\:\mathrm{0}\:\mathrm{and}\:{C}_{\mathrm{2}} \::\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{6}{y}\:+\:\mathrm{9}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:{C}_{\mathrm{3}} \:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centre} \\ $$$$\mathrm{of}\:{C}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:{C}_{\mathrm{1}} \:\mathrm{and}\:{C}_{\mathrm{2}} . \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{two}\:\mathrm{tangents}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{to}\:{C}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{lenght} \\ $$$$\mathrm{of}\:\mathrm{each}\:\mathrm{tangent}. \\ $$
Answered by john santu last updated on 21/Aug/20
(a) C_3 : C_1 +λC_2 =0 ⇒(1+λ)x^2 +(1+λ)y^2 −(6+2λ)x−(4+6λ)y+9+9λ=0 ⇒x^2 +y^2 −(((6+2λ)/(1+λ)))x−(((4+6λ)/(1+λ)))y+((9+9λ)/(1+λ))=0 where { ((((3+2λ)/(1+λ)) = 3⇒3+2λ=3+3λ; λ=0)),((((2+3λ)/(1+λ))= 2 ⇒2+3λ=2+2λ; λ=0)) :} so the equation of C_(3 ) = C_1
$$ \\ $$$$\left({a}\right)\:{C}_{\mathrm{3}} :\:{C}_{\mathrm{1}} +\lambda{C}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+\lambda\right){x}^{\mathrm{2}} +\left(\mathrm{1}+\lambda\right){y}^{\mathrm{2}} −\left(\mathrm{6}+\mathrm{2}\lambda\right){x}−\left(\mathrm{4}+\mathrm{6}\lambda\right){y}+\mathrm{9}+\mathrm{9}\lambda=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\left(\frac{\mathrm{6}+\mathrm{2}\lambda}{\mathrm{1}+\lambda}\right){x}−\left(\frac{\mathrm{4}+\mathrm{6}\lambda}{\mathrm{1}+\lambda}\right){y}+\frac{\mathrm{9}+\mathrm{9}\lambda}{\mathrm{1}+\lambda}=\mathrm{0} \\ $$$${where}\:\begin{cases}{\frac{\mathrm{3}+\mathrm{2}\lambda}{\mathrm{1}+\lambda}\:=\:\mathrm{3}\Rightarrow\mathrm{3}+\mathrm{2}\lambda=\mathrm{3}+\mathrm{3}\lambda;\:\lambda=\mathrm{0}}\\{\frac{\mathrm{2}+\mathrm{3}\lambda}{\mathrm{1}+\lambda}=\:\mathrm{2}\:\Rightarrow\mathrm{2}+\mathrm{3}\lambda=\mathrm{2}+\mathrm{2}\lambda;\:\lambda=\mathrm{0}}\end{cases} \\ $$$${so}\:{the}\:{equation}\:{of}\:{C}_{\mathrm{3}\:} =\:{C}_{\mathrm{1}} \: \\ $$
Commented by Rio Michael last updated on 21/Aug/20
sir fourth and fifth steps not understood.
$$\mathrm{sir}\:\mathrm{fourth}\:\mathrm{and}\:\mathrm{fifth}\:\mathrm{steps}\:\mathrm{not}\:\mathrm{understood}. \\ $$
Commented by Rio Michael last updated on 21/Aug/20
please sir, (b ) part
$$\mathrm{please}\:\mathrm{sir},\:\left(\mathrm{b}\:\right)\:\mathrm{part} \\ $$

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