Circle 4 Question 23
23. Find the equation of the circle which passes through the point $(2,0)$ and whose centre is the limit of the point of intersection of the lines $3 x+5 y=1,(2+c) x+5 c^{2} y=1$ as $c$ tends to 1.
$(1979,3 M)$
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Solution:
- Given lines are
$3 x+5 y-1=0$
and $\quad(2+c) x+5 c^{2} y-1=0$
$\therefore \quad \frac{x}{-5+5 c^{2}}=\frac{y}{-(2+c)+3}=\frac{1}{15 c^{2}-5 c-10}$
$\Rightarrow \quad x=\frac{5\left(c^{2}-1\right)}{5\left(3 c^{2}-c-2\right)}$ and $y=\frac{1-c}{15 c^{2}-5 c-10}$
$\Rightarrow \quad \lim _{c \rightarrow 1} x=\lim _{c \rightarrow 1} \frac{2 c}{6 c-1}$
and
$$ \lim _{c \rightarrow 1} y=\lim _{c \rightarrow 1} \frac{-1}{30 c-5} \Rightarrow \lim _{c \rightarrow 1} x=\frac{2}{5} $$
and
$$ \lim _{c \rightarrow 1} y=-\frac{1}{25} $$
$\therefore$ Centre $=\lim _{c \rightarrow 1} x, \lim _{c \rightarrow 1} y=\frac{2}{5},-\frac{1}{25}$
$\therefore$ Radius $=\sqrt{2-\frac{2}{5}^{2}+0+\frac{1}{25}^{2}}=\sqrt{\frac{64}{25}+\frac{1}{625}}=\frac{\sqrt{1601}}{25}$ $\therefore$ Equation of the circle is $x-\frac{2}{5}^{2}+y+\frac{1}{25}^{2}=\frac{1601}{625}$
$$ \begin{aligned} & \Rightarrow & x^{2}+y^{2}-\frac{4 x}{5}+\frac{2 y}{25}+\frac{4}{25}+\frac{1}{625}-\frac{1601}{625} & =0 \\ & \Rightarrow & 25\left(x^{2}+y^{2}\right)-20 x+2 y-60 & =0 \end{aligned} $$
