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

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(B) Let the centre of the circle be $(h, k)$.Since the circle passes through $(0,0), (x,0),$ and $(0,y)$,the distance from the centre to each point mu

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$\left(\frac{x}{2}, \frac{y}{2}\right)$

Vedclass · Answer

(B) Let the centre of the circle be $(h, k)$.Since the circle passes through $(0,0), (x,0),$ and $(0,y)$,the distance from the centre to each point must be equal to the radius $R$.$h^2 + k^2 = (h-x)^2 + k^2 = h^2 + (k-y)^2$From $h^2 + k^2 = (h-x)^2 + k^2$,we get $h^2 = h^2 - 2hx + x^2$,which implies $2hx = x^2$,so $h = \frac{x}{2}$.From $h^2 + k^2 = h^2 + (k-y)^2$,we get $k^2 = k^2 - 2ky + y^2$,which implies $2ky = y^2$,so $k = \frac{y}{2}$.Thus,the coordinates of the centre are $\left(\frac{x}{2}, \frac{y}{2}\right)$.

If a circle passes through the points $(0,0), (x,0),$ and $(0,y)$,then the coordinates of its centre are

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