Question

Find the equations of a circle which passes though the point ( 2 , 0 ) and whose center is the limit of the point of intersection of the lines 3x + 5y = 1 , (2 + c ) x + $5c^2$ y = 1 tends to 1

Answer 1

Eliminating y, we get
(3c${}^2$2 c - 2)x = c${}^2$ - 1 $\therefore x=\frac{(c-1)(c+1)}{(c-1)(3c+2)}=\frac{c+1}{3c+2}$
$\therefore$ limit when c $\Rightarrow$ 1 gives $x=\frac{2}{5}$ and hence $y=-\frac{1}{25}$.
$\therefore$ Centre ( g , f) is $(\frac{2}{5},\frac{1}{25})$.
Circle is
$x^2+y^2-\left(\frac{4}{5}\right)x+\left(\frac{2}{25}\right)y+k=0$
It passes through $(2,0)$ $\therefore$ $k=-\left(\frac{12}{5}\right)$
Hence the required circle is
$25(x^2+y^2)-20x+60y=0.$