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  • The limiting points of the coaxial system containing the two circles <img alt="\mathrm{x^2+y^2+x-y+1=0 \text { and } 5\left(x^2+y^2\right)-2 x-16 y+13=0}" src="https://entrancecorner.oncodecogs.com

The limiting points of the coaxial system containing the two circles \mathrm{x^2+y^2+x-y+1=0 \text { and } 5\left(x^2+y^2\right)-2 x-16 y+13=0} are

Option: 1

(1,-1),(-5,-40)


Option: 2

(1,-1),(-1 / 5,-8 / 5)


Option: 3

\left(\frac{-12}{17}, \frac{3}{17}\right),(1 / 5,8 / 5)


Option: 4

(-1,1),(-1 / 5,-8 / 5)


Answers (1)

best_answer

The equation of the radical axis of the given circles is 7x + 11y – 8 = 0. Therefore, the equation of family of coaxial circles is

\mathrm{\begin{aligned} & x^2+y^2+x-y+1+\lambda(7 x+11 y-8)=0 \\ \Rightarrow & x^2+y^2+x(1+7 \lambda)+y(11 \lambda-1)+1-8 \lambda=0 \end{aligned}}

The coordinates of the centre and radius of this circle are 

\mathrm{\begin{aligned} & -\left(\frac{7 \lambda+1}{2}\right),-\left(\frac{11 \lambda-1}{2}\right) \\ & \text { and } \sqrt{\left(\frac{7 \lambda+1}{2}\right)^2+\left(\frac{11 \lambda-1}{2}\right)^2-(1-8 \lambda)} \end{aligned}}

For limiting points, we must have
Radius = 0

\mathrm{\begin{aligned} & \Rightarrow\left(\frac{7 \lambda+1}{2}\right)^2+\left(\frac{11 \lambda-1}{2}\right)^2-(1-8 \lambda)=0 \\ & \Rightarrow(5 \lambda+1)(17 \lambda-1)=0 \\ & \Rightarrow \lambda=\frac{-1}{5}, \frac{1}{17} \end{aligned}}

\mathrm{\text { Hence, the limiting points are }\left(\frac{-12}{17}, \frac{3}{17}\right) \text { and }(1 / 5,8 / 5) \text {. }}

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