For any real number lambda neq 1,the centre o | vedclass

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(A) Let the circle pass through $A(1, \lambda)$,$B(\lambda, 1)$,and $C(\lambda, \lambda)$.Since the points $B(\lambda, 1)$ and $C(\lambda, \lambda)$ h

Vedclass · Accepted Answer

$\left(\frac{1+\lambda}{2}, \frac{1+\lambda}{2}\right)$

Vedclass · Answer

(A) Let the circle pass through $A(1, \lambda)$,$B(\lambda, 1)$,and $C(\lambda, \lambda)$.Since the points $B(\lambda, 1)$ and $C(\lambda, \lambda)$ have the same $x$-coordinate,the perpendicular bisector of $BC$ is the horizontal line $y = \frac{1+\lambda}{2}$.Since the points $A(1, \lambda)$ and $C(\lambda, \lambda)$ have the same $y$-coordinate,the perpendicular bisector of $AC$ is the vertical line $x = \frac{1+\lambda}{2}$.The centre of the circle is the intersection of these perpendicular bisectors.Therefore,the centre is $\left(\frac{1+\lambda}{2}, \frac{1+\lambda}{2}\right)$.

For any real number $\lambda \neq 1$,the centre of the circle that passes through $A(1, \lambda)$,$B(\lambda, 1)$,and $C(\lambda, \lambda)$ is

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