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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
$( - 1,\,1)$
$( - 1,\,2)$
$( - 2,\,1)$
$( - 2,\,2)$
Solution
(a) Trick : The equation of radical axis is ${S_1} – {S_2} = 0$
$i.e.$, $4x + 2y – 1 = 0$.
$\therefore $ 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 ${x^2} + {y^2} – (6 – 4\lambda )x – (6 – 2\lambda )y + (4 – \lambda ) = 0$….(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 $\left( {\frac{1}{5},\;\frac{8}{5}} \right)$
One of these limit points is given in $(a).$
