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  • If (-2,1)is a limitting point of a co-axial syatem of circle of which <img alt="\mathrm{x^2+y^2-4 x-6 y+7=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%

If (-2,1)is a limitting point of a co-axial syatem of circle of which \mathrm{x^2+y^2-4 x-6 y+7=0}  is a member, then the other limiting point is 

Option: 1

\mathrm{\left(\frac{4}{5}, \frac{-12}{5}\right)}


Option: 2

\mathrm{\left(\frac{-4}{5}, \frac{12}{5}\right)}


Option: 3

\mathrm{\left(\frac{-1}{5}, \frac{-12}{5}\right)}


Option: 4

\mathrm{\left(\frac{4}{5}, \frac{12}{5}\right)}


Answers (1)

best_answer

 As one of the limiting point of co-axial system is (–2, 1).
\therefore Point circle is \mathrm{(x+2)^2+(y-1)^2=0}

Or \mathrm{x^2+y^2+4 x-2 y+5=0}

\mathrm{\begin{aligned} & \Rightarrow(1+\lambda)\left(x^2+y^2\right)-2(2-2 \lambda) x-2(3+\lambda) y+(5 \lambda+7)=0 \\ & \Rightarrow x^2+y^2-\frac{2(2-2 \lambda)}{1+\lambda} x-\frac{2(3+\lambda)}{1+\lambda} y+\frac{(5 \lambda+7)}{1+\lambda}=0 \end{aligned}}

\mathrm{\therefore \quad \text { Centre is }\left(\frac{2-2 \lambda}{1+\lambda}, \frac{3+\lambda}{1+\lambda}\right)} ----(i)

And radius \mathrm{=0 \Rightarrow\left(\frac{2-2 \lambda}{1+\lambda}\right)^2+\left(\frac{3+\lambda}{1+\lambda}\right)^2-\frac{(5 \lambda+7)}{1+\lambda}=0}

\mathrm{\begin{aligned} & \Rightarrow(2-2 \lambda)^2+(3+\lambda)^2-(5 \lambda+7)(\lambda+1)=0 \\ & \Rightarrow-14 \lambda+6=0 \quad \therefore \lambda=\frac{3}{7} \end{aligned}}

\mathrm{\therefore } from (1) , the other limiting point is \mathrm{\left(2\left(\frac{1-\lambda}{1+\lambda}\right) \cdot \frac{3+\lambda}{1+\lambda}\right) }

 

\mathrm{=\left(2 \cdot \frac{4}{7} \times \frac{7}{10}, \frac{3+\frac{3}{7}}{\frac{10}{7}}\right)=\left(\frac{4}{5}, \frac{12}{5}\right) }

 

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