If a circle passes through points (4,0) and ( | vedclass

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(B) Let the centre of the circle be $(0, y)$.Since the circle passes through $(4, 0)$ and $(0, 2)$,the distance from the centre to these points must b

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$21$

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(B) Let the centre of the circle be $(0, y)$.Since the circle passes through $(4, 0)$ and $(0, 2)$,the distance from the centre to these points must be equal to the radius $r$.$\sqrt{(4-0)^2 + (0-y)^2} = \sqrt{(0-0)^2 + (2-y)^2}$$16 + y^2 = (2-y)^2$$16 + y^2 = 4 - 4y + y^2$$16 = 4 - 4y$$4y = -12$$y = -3$Thus,the centre is $(0, -3)$.The radius $r$ is the distance from $(0, -3)$ to $(0, 2)$:$r = \sqrt{(0-0)^2 + (2 - (-3))^2} = \sqrt{0^2 + 5^2} = 5$.Now,calculate $r^2 - r + 1$:$r^2 - r + 1 = 5^2 - 5 + 1 = 25 - 5 + 1 = 21$.

If a circle passes through points $(4,0)$ and $(0,2)$ and its centre lies on the $Y$-axis. If the radius of the circle is $r$,then the value of $r^2-r+1$ is

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