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  • The point is a limiting point of a co-axial system of circles of which <img alt="\mathrm{x^2+y^2=9}" src="https

The point (2,3) is a limiting point of a co-axial system of circles of which \mathrm{x^2+y^2=9} is a member. The coordinates of the other limiting point is given by

Option: 1

\left(\frac{18}{13}, \frac{27}{13}\right)


Option: 2

\left(\frac{9}{13}, \frac{6}{13}\right)


Option: 3

\left(\frac{18}{13},-\frac{27}{13}\right)


Option: 4

\left(-\frac{18}{13},-\frac{9}{13}\right)


Answers (1)

Equation of circle with (2, 3) as limiting point is \mathrm{(x-2)^2+(y-3)^2=0}

or \mathrm{\left(x^2+y^2-9\right)-4 x-6 y+22=0} or \mathrm{\left(x^2+y^2-9\right)-\lambda(2 x+3 y-11)=0 \quad} represents the family of co-axial circles.

\mathrm{ \begin{aligned} & c=\left(\lambda, \frac{3 \lambda}{2}\right), r=\sqrt{\lambda^2+\frac{9 \lambda^2}{4}-11 \lambda+9} \quad \text { For limiting points } r=0 \\ & \Rightarrow 13 \lambda^2-44 \lambda+36=0 \Rightarrow \lambda=\frac{18}{13}, 2 \\ & \therefore \text { The limiting points are }(2,3) \text { and }\left[\frac{18}{13}, \frac{3}{2}\left(\frac{18}{13}\right)\right] \text { or }\left(\frac{18}{13}, \frac{27}{13}\right) \end{aligned} }

Posted by

Ramraj Saini

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