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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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