find the equation of circle passing through origin of radius 5 and x- intercept substend anangle of 60 degree at the centre

Dear Student,
Please find below the solution to the asked query:


$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{OR}\hspace{0.17em}\mathrm{subtends} 60° \mathrm{at} \mathrm{centre}. \\ \left(\mathrm{a},\mathrm{b}\right) \mathrm{are} \mathrm{coordinates} \mathrm{of} \mathrm{centre} \\ \mathrm{Now} \angle \mathrm{OPQ}=30° \\ \mathrm{r}=\mathrm{OP}=5 \\ \mathrm{OQ}=\left|\mathrm{a}\right| \\ \sin 30°=\frac{\left|\mathrm{a}\right|}{5} \\ \Rightarrow \frac{1}{2}=\frac{\left|\mathrm{a}\right|}{5} \\ \Rightarrow \left|\mathrm{a}\right|=\frac{5}{2} \\ \Rightarrow \mathrm{a}=\pm \frac{5}{2} \\ \cos 30°=\frac{\left|\mathrm{b}\right|}{5} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\left|\mathrm{b}\right|}{5} \\ \Rightarrow \left|\mathrm{b}\right|=\frac{5\sqrt{3}}{2} \\ \Rightarrow \mathrm{b}=\pm \frac{5\sqrt{3}}{2} \\ \mathrm{Hence} \mathrm{equations} \mathrm{will} \mathrm{be} \\ {\left(\mathrm{x}-\frac{5}{2}\right)}^2+{\left(\mathrm{x}-\frac{5\sqrt{3}}{2}\right)}^2=25 \mathrm{or} \\ {\left(\mathrm{x}+\frac{5}{2}\right)}^2+{\left(\mathrm{x}-\frac{5\sqrt{3}}{2}\right)}^2=25 \mathrm{or} \\ {\left(\mathrm{x}-\frac{5}{2}\right)}^2+{\left(\mathrm{x}+\frac{5\sqrt{3}}{2}\right)}^2=25 \mathrm{or} \\ {\left(\mathrm{x}+\frac{5}{2}\right)}^2+{\left(\mathrm{x}+\frac{5\sqrt{3}}{2}\right)}^2=25 \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards