One of the limit point of the coaxial system | vedclass

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Question

(a) Trick : The equation of radical axis is ${S_1} - {S_2} = 0$

$i.e.$, $4x + 2y - 1 = 0$.

$	herefore $ The equation of circle of co-axial syst

Vedclass · Accepted Answer

$( - 1,\,1)$

Vedclass · Answer

(a) Trick : The equation of radical axis is ${S_1} - {S_2} = 0$

$i.e.$, $4x + 2y - 1 = 0$.

$	herefore $ The equation of circle of co-axial system can be taken as $({x^2} + {y^2} - 6x - 6y + 4) + \lambda (4x + 2y - 1) = 0$

or ${x^2} + {y^2} - (6 - 4\lambda )x - (6 - 2\lambda )y + (4 - \lambda ) = 0$&hellip;.(i)

whose centre is $C(3 - 2\lambda ,\;3 - \lambda )$ and radius is

$r = \sqrt {{{(3 - 2\lambda )}^2} + {{(3 - \lambda )}^2} - (4 - \lambda )} $

If $r = 0$, then we get $\lambda = 2$ or $7/5$.

Putting the co-ordinates of $C$, the limit points are $(-1, 1)$ and $\left( {\frac{1}{5},\;\frac{8}{5}} ight)$

One of these limit points is given in $(a).$

One of the limit point of the coaxial system of circles containing ${x^2} + {y^2} - 6x - 6y + 4 = 0$, ${x^2} + {y^2} - 2x$ $ - 4y + 3 = 0$ is

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