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Question-38289




Question Number 38289 by Tinkutara last updated on 23/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18
centre(−g,−f) lies on ax+by+d=0  a(−g)+b(−f)+d=0   λ(−g)−a(−f)+μg+k=0  g(−a)+f(−b)+d=0  g(−λ+μ)+f(a)+k=0  g(−a)+f(−b)=−d  (g/(d/a))+(f/(d/b))=1  (g/(k/(λ−μ)))+(f/(k/a))=1  (d/a)=(k/(λ−μ))    and (d/b)=(k/a)  (d/a)=((ad)/(b(λ−μ)))    b(λ−μ)=a^2   λ−μ=(a^2 /b)  pls check
$${centre}\left(−{g},−{f}\right)\:{lies}\:{on}\:{ax}+{by}+{d}=\mathrm{0} \\ $$$${a}\left(−{g}\right)+{b}\left(−{f}\right)+{d}=\mathrm{0}\: \\ $$$$\lambda\left(−{g}\right)−{a}\left(−{f}\right)+\mu{g}+{k}=\mathrm{0} \\ $$$${g}\left(−{a}\right)+{f}\left(−{b}\right)+{d}=\mathrm{0} \\ $$$${g}\left(−\lambda+\mu\right)+{f}\left({a}\right)+{k}=\mathrm{0} \\ $$$${g}\left(−{a}\right)+{f}\left(−{b}\right)=−{d} \\ $$$$\frac{{g}}{\frac{{d}}{{a}}}+\frac{{f}}{\frac{{d}}{{b}}}=\mathrm{1} \\ $$$$\frac{{g}}{\frac{{k}}{\lambda−\mu}}+\frac{{f}}{\frac{{k}}{{a}}}=\mathrm{1} \\ $$$$\frac{{d}}{{a}}=\frac{{k}}{\lambda−\mu}\:\:\:\:{and}\:\frac{{d}}{{b}}=\frac{{k}}{{a}} \\ $$$$\frac{{d}}{{a}}=\frac{{ad}}{{b}\left(\lambda−\mu\right)} \\ $$$$ \\ $$$${b}\left(\lambda−\mu\right)={a}^{\mathrm{2}} \\ $$$$\lambda−\mu=\frac{{a}^{\mathrm{2}} }{{b}}\:\:{pls}\:{check} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 23/Jun/18
λ+μ is to be found, Sir.
$$\lambda+\mu\:{is}\:{to}\:{be}\:{found},\:{Sir}. \\ $$

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