For any real number $\lambda \neq 1$, the centre of the circle that passes through $\mathrm{A}(1, \lambda),…

For any real number $\lambda \neq 1$, the centre of the circle that passes through $\mathrm{A}(1, \lambda), \mathrm{B}(\lambda, 1)$ and $(\lambda, \lambda)$ is
  1. $\left(\frac{1+\lambda}{2}, \frac{1+\lambda}{2}\right)$
  2. $\left(\frac{1+2 \lambda}{3}, \frac{1+2 \lambda}{3}\right)$
  3. $(1+2 \lambda, 1+2 \lambda)$
  4. $\left(\frac{\lambda}{2}, \frac{\lambda}{2}\right)$

Solution

Let circles passes through $\mathrm{A}(1, \lambda),(\mathrm{BC} \lambda, 1)$ and $\mathrm{C}(\lambda, \lambda)$ hence, equation of circle are $\begin{aligned} & 1+\lambda^2+2 \mathrm{~g}+2+\lambda+\mathrm{c}=0 \\ & \lambda^2+1+2 \mathrm{~g} \lambda+2 \mathrm{f}+\mathrm{c}=0 \\ & \lambda^2+\lambda^2+2 \mathrm{~g} \lambda+2 \mathrm{f} \lambda+\mathrm{c}=0\end{aligned}$ on solving equations, $\mathrm{f}=\mathrm{g}$ and $\mathrm{f}=\frac{-\lambda-1}{2}$ how centre $=(-\mathrm{g},-\mathrm{f})=\left(\frac{1+\lambda}{2}, \frac{1+\lambda}{2}\right)$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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