Let a circle passing through (2,0) have its centre at the point [(h, k)].

Question

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 :

Accepted Answer

$$\begin{aligned}
& 3 x+5 y=1 \
& (2+c) x+5 c^2 y=1 \
& 3 c^2 x+5 c^2 y=c^2
\end{aligned}$$

Subtracting

$$\begin{aligned}
& \left(2+c-3 c^2\right) x=1-c^2 \
& x_c=\frac{1-c^2}{2+c-3 c^2}=\frac{(1-c)(1+c)}{(1-c)(3 c+2)}=\frac{c+1}{3 c+2} \
& y=\frac{1-3 x}{5}=\frac{1-3\left(\frac{c+1}{3 c+2}\right)}{5} \
& =\frac{3 c+2-3 c-3}{5(3 c+2)} \
& y_c=\frac{-1}{5(3 c+2)} \
& \lim _{c \rightarrow 1} x_c=\frac{2}{5}=h \
& \lim _{c \rightarrow 1} y_c=\frac{-1}{25}=k
\end{aligned}$$

Equation of circle is

$25 x^2+25 y^2-20 x+2 y-60=0$

Solutions · 1