The lengths of the intercepts made by a circle $\mathrm{S}$ on $\mathrm{X}$ and $\mathrm{Y}$ - axes are…
The lengths of the intercepts made by a circle $\mathrm{S}$ on $\mathrm{X}$ and $\mathrm{Y}$ - axes are $\frac{2 \sqrt{13}}{3}$ and $\frac{2 \sqrt{22}}{3}$ respectively. If the radius of the circle $\mathrm{S}$ is $\frac{\sqrt{38}}{3}$ and its centre $\mathrm{C}$ lies in the second quadrant, then $\mathrm{C}=$
$\left(\frac{-5}{3}, \frac{4}{3}\right)$
$\left(\frac{-4}{3}, \frac{5}{3}\right)$
$\left(\frac{-6}{5}, \frac{7}{5}\right)$
$\left(\frac{-7}{5}, \frac{6}{5}\right)$
Solution
Let equation of the circle be
$x^2+y^2+2 g x+2 f y+c=0$ ... (i)
Now length of intercepts made by $\mathrm{x}$ axis is
$2 \sqrt{\mathrm{g}^2-\mathrm{c}}=\frac{2 \sqrt{13}}{3} \Rightarrow \mathrm{g}^2-\mathrm{c}=\frac{13}{9}$ ... (ii)
Similarly, $2 \sqrt{\mathrm{f}^2-\mathrm{c}}=\frac{2 \sqrt{22}}{3} \Rightarrow \mathrm{f}^2-\mathrm{c}=\frac{22}{9}$ ... (iii)
$\text { Since, radius }=\frac{\sqrt{38}}{3}$
$\begin{aligned}
& \Rightarrow \sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\frac{\sqrt{38}}{3} \Rightarrow \mathrm{g}^2+\mathrm{f}^2-\mathrm{c}=\frac{38}{9} \\
& \Rightarrow \frac{13}{9}+\frac{22}{9}+\mathrm{C}=\frac{38}{9} \quad \text { (by (ii) & (iii)) } \\
& \Rightarrow \mathrm{C}=\frac{3}{9}=\frac{1}{3}
\end{aligned}$
So $\mathrm{g}^2=\frac{13}{9}+\frac{1}{3}=\frac{16}{9} \Rightarrow \mathrm{g}= \pm \frac{4}{3}$
$\& \mathrm{f}^2=\frac{22}{9}+\frac{1}{3}=\frac{25}{9} \Rightarrow \mathrm{f}= \pm \frac{5}{3}$
Since centre lies in 2nd quadrant so centre
$=\left(-\frac{4}{3}, \frac{5}{3}\right)$