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

$$\mathrm{Given}\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{twe}\mathrm{circles}$$$$C_1:x^2+y^2-6x-4y+9=0\mathrm{and}{C}_2:x^2+y^2-2x-6y+9.$$$$\left(\mathrm{a}\right)\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{circle}{C}_3\mathrm{which}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{centre}$$$$\mathrm{of}{C}_1\mathrm{and}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}{C}_1\mathrm{and}{C}_2.$$$$\left(\mathrm{b}\right)\mathrm{The}\mathrm{equations}\mathrm{of}\mathrm{two}\mathrm{tangents}\mathrm{from}\mathrm{the}\mathrm{origin}\mathrm{to}{C}_1\mathrm{and}\mathrm{the}\mathrm{lenght}$$$$\mathrm{of}\mathrm{each}\mathrm{tangent}.$$

Answered by john santu last updated on 21/Aug/20

$$$$$$\left(a\right)C_3:C_1+\lambda C_2=0$$$$\Rightarrow (1+\lambda )x^2+(1+\lambda )y^2-(6+2\lambda )x-(4+6\lambda )y+9+9\lambda =0$$$$\Rightarrow x^2+y^2-\left(\frac{6+2\lambda }{1+\lambda }\right)x-\left(\frac{4+6\lambda }{1+\lambda }\right)y+\frac{9+9\lambda }{1+\lambda }=0$$$$where\{\begin{array}{l}\frac{3+2\lambda }{1+\lambda }=3\Rightarrow 3+2\lambda =3+3\lambda ;\lambda =0\\ \frac{2+3\lambda }{1+\lambda }=2\Rightarrow 2+3\lambda =2+2\lambda ;\lambda =0\end{array}$$$$sotheequationof{C}_3=C_1$$

Commented by Rio Michael last updated on 21/Aug/20

$$\mathrm{sir}\mathrm{fourth}\mathrm{and}\mathrm{fifth}\mathrm{steps}\mathrm{not}\mathrm{understood}.$$

Commented by Rio Michael last updated on 21/Aug/20

$$\mathrm{please}\mathrm{sir},\left(\mathrm{b}\right)\mathrm{part}$$

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