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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$ The length of the intercept made by the circle on the $X$-axis is $2\sqrt{g^2-c}

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$\left(\frac{-4}{3}, \frac{5}{3}\right)$

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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$ The length of the intercept made by the circle on the $X$-axis is $2\sqrt{g^2-c} = \frac{2\sqrt{13}}{3} \Rightarrow g^2-c = \frac{13}{9}$ ... $(ii)$ The length of the intercept made by the circle on the $Y$-axis is $2\sqrt{f^2-c} = \frac{2\sqrt{22}}{3} \Rightarrow f^2-c = \frac{22}{9}$ ... $(iii)$ Given the radius $r = \frac{\sqrt{38}}{3}$,we have $r^2 = g^2+f^2-c = \frac{38}{9}$ ... $(iv)$ From $(ii)$,$g^2 = c + \frac{13}{9}$. From $(iii)$,$f^2 = c + \frac{22}{9}$. Substituting these into $(iv)$: $(c + \frac{13}{9}) + (c + \frac{22}{9}) - c = \frac{38}{9}$ $c + \frac{35}{9} = \frac{38}{9} \Rightarrow c = \frac{3}{9} = \frac{1}{3}$ Now,$g^2 = \frac{1}{3} + \frac{13}{9} = \frac{16}{9} \Rightarrow g = \pm \frac{4}{3}$ And $f^2 = \frac{1}{3} + \frac{22}{9} = \frac{25}{9} \Rightarrow f = \pm \frac{5}{3}$ Since the centre $C(-g, -f)$ lies in the second quadrant,$g$ must be positive and $f$ must be negative. Thus,$C = (-\frac{4}{3}, \frac{5}{3})$.

The lengths of the intercepts made by a circle $S$ on $X$ and $Y$-axes are $\frac{2 \sqrt{13}}{3}$ and $\frac{2 \sqrt{22}}{3}$ respectively. If the radius of the circle $S$ is $\frac{\sqrt{38}}{3}$ and its centre $C$ lies in the second quadrant,then $C=$

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