JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 10)
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 :
$$5 x^2+5 y^2-4 x-2 y-12=0$$
$$25 x^2+25 y^2-20 x+2 y-60=0$$
$$25 x^2+25 y^2-2 x+2 y-60=0$$
$$5 x^2+5 y^2-4 x+2 y-12=0$$
Explanation
$$\begin{aligned} & 3 x+5 y=1 \\ & (2+c) x+5 c^2 y=1 \\ & 3 c^2 x+5 c^2 y=c^2 \end{aligned}$$
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
$$25 x^2+25 y^2-20 x+2 y-60=0$$
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