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

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(0, 0)$,we have $0^2 + 0^2 + 2g(0) + 2f(0) + c

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$\left( \frac{a}{2}, \frac{b}{2} \right)$

Vedclass · Answer

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(0, 0)$,we have $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$,which gives $c = 0$.Since it passes through $(a, 0)$,we have $a^2 + 0^2 + 2g(a) + 2f(0) + 0 = 0$,which implies $a^2 + 2ga = 0$,so $g = -\frac{a}{2}$.Since it passes through $(0, b)$,we have $0^2 + b^2 + 2g(0) + 2f(b) + 0 = 0$,which implies $b^2 + 2fb = 0$,so $f = -\frac{b}{2}$.The centre of the circle is $(-g, -f)$.Substituting the values of $g$ and $f$,the centre is $\left( -(-\frac{a}{2}), -(-\frac{b}{2}) \right) = \left( \frac{a}{2}, \frac{b}{2} \right)$.

If a circle passes through the points $(0, 0)$,$(a, 0)$,and $(0, b)$,then its centre is:

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