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(C) The centre of the circle is $(\alpha, \beta)$ and it passes through the origin $(0, 0)$.The radius $r$ is the distance between the centre $(\alpha

Vedclass · Accepted Answer

$x^2 + y^2 - 2\alpha x - 2\beta y = 0$

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(C) The centre of the circle is $(\alpha, \beta)$ and it passes through the origin $(0, 0)$.The radius $r$ is the distance between the centre $(\alpha, \beta)$ and the origin $(0, 0)$:$r = \sqrt{(\alpha - 0)^2 + (\beta - 0)^2} = \sqrt{\alpha^2 + \beta^2}$.The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = \alpha$,$k = \beta$,and $r^2 = \alpha^2 + \beta^2$:$(x - \alpha)^2 + (y - \beta)^2 = \alpha^2 + \beta^2$Expanding the squares:$x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2 = \alpha^2 + \beta^2$Simplifying the equation:$x^2 + y^2 - 2\alpha x - 2\beta y = 0$.

If $(\alpha, \beta)$ is the centre of a circle passing through the origin,then its equation is

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