Let a circle passing through $(2,0)$ have its centre at the point $(\mathrm{h}, \mathrm{k})$. Let $(x_{\mathrm{c}}, y_{\mathrm{c}})$ be the point of intersection of the lines $3 x+5 y=1$ and $(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1$. If $\mathrm{h}=\lim _\limits{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}$ and $\mathrm{k}=\lim _\limits{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}$, then the equation of the circle is :
Solution
<p>$$\begin{aligned}
& 3 x+5 y=1 \\
& (2+c) x+5 c^2 y=1 \\
& 3 c^2 x+5 c^2 y=c^2
\end{aligned}$$</p>
<p>Subtracting</p>
<p>$$\begin{aligned}
& \left(2+c-3 c^2\right) x=1-c^2 \\
& x_c=\frac{1-c^2}{2+c-3 c^2}=\frac{(1-c)(1+c)}{(1-c)(3 c+2)}=\frac{c+1}{3 c+2} \\
& y=\frac{1-3 x}{5}=\frac{1-3\left(\frac{c+1}{3 c+2}\right)}{5} \\
& =\frac{3 c+2-3 c-3}{5(3 c+2)} \\
& y_c=\frac{-1}{5(3 c+2)} \\
& \lim _{c \rightarrow 1} x_c=\frac{2}{5}=h \\
& \lim _{c \rightarrow 1} y_c=\frac{-1}{25}=k
\end{aligned}$$</p>
<p>Equation of circle is</p>
<p>$25 x^2+25 y^2-20 x+2 y-60=0$</p>
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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