Find the equation of circle whose centre is (3,-1) and which cut off an intercept of length 6 units from the line 2x-5y+18=0.

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$$\begin{gathered} \mathrm{We} \mathrm{have}, \mathrm{centre} \mathrm{of} \mathrm{circle} \mathrm{C}\left(\mathrm{h},\mathrm{k}\right) = \mathrm{C}\left(3,-1\right) \\ \mathrm{equation} \mathrm{of} \mathrm{chord} , \mathrm{AB}=2\mathrm{x}-5\mathrm{y}+18 \\ \mathrm{Now}, \mathrm{we} \mathrm{drop} \mathrm{a} \mathrm{perpendicular} \mathrm{CM} \mathrm{on} \mathrm{the} \mathrm{chord}, \\ \mathrm{CM}=\frac{2.3-5.\left(-1\right)+18}{\sqrt{{\left(2\right)}^2+{\left(-5\right)}^2}}=\frac{29}{\sqrt{29}}=\sqrt{29} \mathrm{units} \\ \mathrm{MB}=3 \mathrm{units} \left(\perp \mathrm{drawn} \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{to} \mathrm{the} \mathrm{chord} \mathrm{bisects} \mathrm{the} \mathrm{chord}\right) \\ \mathrm{In} ∆\mathrm{CMB}, \\ {\mathrm{CB}}^2={\mathrm{CM}}^2+{\mathrm{MB}}^2 \left(\mathrm{Pythagoras} \mathrm{Theorem}\right) \\ {\mathrm{CB}}^2={\left(\sqrt{29}\right)}^2+{\left(3\right)}^2 \\ {\mathrm{CB}}^2=29+9=38 \\ \mathrm{CB}=\sqrt{38}\mathrm{units} \\ \mathrm{r}=\mathrm{CB} \\ \mathbf{hence}\mathbf{ }\mathbf{equation}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{circle}\mathbf{:} \\ {\left(\mathbf{x}\mathbf{-}\mathbf{h}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\left(\mathbf{y}\mathbf{-}\mathbf{k}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{r}}^{\mathbf{2}} \\ {\left(\mathbf{x}\mathbf{-}\mathbf{3}\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathbf{y}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{38} \end{gathered}$$

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