Let a circle passing through $(2,0)$ have its centre at the point $(h, k)$. Let $\left(x_c, y_c\right)$ be…

Let a circle passing through $(2,0)$ have its centre at the point $(h, k)$. Let $\left(x_c, y_c\right)$ be the point of intersection of the lines $3 x+5 y=1$ and $(2+c) x+5 c^2 y=1$. If $\mathrm{h}=\lim _{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}$ and $\mathrm{k}=\lim _{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}$, then the equation of the circle is :
  1. $25 x^2+25 y^2-2 x+2 y-60=0$
  2. $5 x^2+5 y^2-4 x+2 y-12=0$
  3. $5 x^2+5 y^2-4 x-2 y-12=0$
  4. $25 x^2+25 y^2-20 x+2 y-60=0$

Solution

$\begin{aligned} & (2+c) x+5 c^2\left(\frac{1-3 x}{5}\right)=1 \\ & x=\frac{1-c^2}{2+c-3 c^2}, y=\frac{1-3 x}{5}=\frac{c-1}{5\left(2+c-3 c^2\right)} \\ & h=\lim _{c \rightarrow 1} \frac{(1-c)(1+c)}{(1-c)(2+3 c)}=\frac{2}{5} \\ & K=\lim _{c \rightarrow 1} \frac{c-1}{-5(c-1)(3 c+2)}=-\frac{1}{25} \\ & \operatorname{Centre}\left(\frac{2}{25},-\frac{1}{25}\right)\end{aligned}$ $\begin{aligned} & r=\sqrt{\left(2-\frac{2}{5}\right)^2+\left(0-\frac{1}{25}\right)^2}=\sqrt{\frac{64}{25}+\frac{1}{625}} \\ & r=\frac{\sqrt{161}}{25} \\ & \left(x-\frac{2}{5}\right)^2+\left(y+\frac{1}{25}\right)^2=\frac{161}{125} \\ & \Rightarrow 25 x^2+25 y^2-20 x+2 y-60=0\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 1)

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