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  • The co-ordinates of the limiting points of the co-axial system of circles <img alt="\mathrm{x^2+y^2+3 x+2 y-4=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plu

The co-ordinates of the limiting points of the co-axial system of circles \mathrm{x^2+y^2+3 x+2 y-4=0} and \mathrm{x^2+y^2-x-2 y+4=0} are 

Option: 1

\mathrm{\left(\frac{3-\sqrt{23}}{4}, \frac{5-\sqrt{23}}{4}\right)}


Option: 2

\mathrm{\left(\frac{3+\sqrt{23}}{4}, \frac{5+\sqrt{23}}{4}\right)}


Option: 3

Both (a) and (b) 


Option: 4

None of these


Answers (1)

best_answer

\therefore Radical axis of the given circles is \mathrm{S_1-S_2=0}

\mathrm{\begin{aligned} & \Rightarrow\left(x^2+y^2+3 x+2 y-4\right)-\left(x^2+y^2-x-2 y+4\right)=0 \\ & \Rightarrow 4 x+4 y-8=0 \Rightarrow x+y-2=0=L \end{aligned}}

Now, system of co-axial circles is given by

\mathrm{\begin{aligned} & S_1+\lambda I=0=S_2+\lambda I \\ \Rightarrow & \left(x^2+y^2+3 x+2 y-4\right)+\lambda(x+y-2)=0 \\ \Rightarrow & x^2+y^2+2 x\left(\frac{3}{2}+\frac{\lambda}{2}\right)+2 y\left(1+\frac{\lambda}{2}\right)+(-4-2 \lambda)=0 \end{aligned}} ----(1)

\mathrm{\rightarrow } Centre of co-axial system of circles is \mathrm{\left(-\frac{3}{2}-\frac{\lambda}{2},-1-\frac{\lambda}{2}\right) }

Now to determine the value of \mathrm{\lambda }, putting R = 0 in equation

\mathrm{\begin{aligned} & \text { (i), } R=\sqrt{G^2+F^2-C}=0 \Rightarrow G^2+F^2-C=0 \\ & \Rightarrow\left(-\frac{3}{2}-\frac{\lambda}{2}\right)^2+\left(-1-\frac{\lambda}{2}\right)^2-(-4-2 \lambda)=0 \\ & \Rightarrow(3+\lambda)^2+(2+\lambda)^2+4(4+2 \lambda)=0 \\ & \Rightarrow 2 \lambda^2+18 \lambda+29=0 \\ & \Rightarrow \lambda=\frac{-9 \pm \sqrt{23}}{2} \therefore \lambda=\frac{-9+\sqrt{23}}{2}, \frac{-9-\sqrt{23}}{2} \end{aligned} }

Putting value of \mathrm{\lambda} in \mathrm{-\frac{1}{2}(3+\lambda)} and \mathrm{-\frac{1}{2}(2+\lambda)} we have

\mathrm{\begin{aligned} & \left(-\frac{1}{2}\left(3+\left(-\frac{9}{2}\right)+\frac{\sqrt{23}}{2}\right),-\frac{1}{2}\left(2-\frac{9}{2}+\frac{\sqrt{23}}{2}\right)\right) \\ & =\left(-\frac{1}{4}(-3+\sqrt{23}),-\frac{1}{4}(-5+\sqrt{23})\right) \end{aligned}}

and

 \mathrm{\begin{aligned} & \left(-\frac{1}{2}\left(3-\frac{9}{2}-\frac{\sqrt{23}}{2}\right),-\frac{1}{2}\left(2-\frac{9}{2}-\frac{\sqrt{23}}{2}\right)\right) \\ & =\left(\frac{1}{4}(3+\sqrt{23}) \cdot \frac{1}{4}(5+\sqrt{23})\right), \end{aligned}}

which are required limiting points.

 

Posted by

Ritika Jonwal

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