Find the mid point of the chord intercepted by the circle x² + y² - 2x - 10 y + 1 = 0 on the line x – 2y + 7 = 0.

$\text{ Given circle is }{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-10\mathrm{y}+1=0$
$\text{ Centre }\mathrm{C}=(-\mathrm{g},-\mathrm{f})=(1,5)$

Equation of the chord is x - 2y + 7 = 0 .
$(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\text{ form })$
Let P(h, k) be the
$\text{ Then }\mathrm{P}(\mathrm{h},\mathrm{k})=\text{ Foot of the }{\perp }^{\text{er }}\text{ from }$
$\mathrm{C}\left({\mathrm{x}}_1{\mathrm{y}}_1\right)=(1,5)\text{ to the line }\mathrm{x}-2\mathrm{y}+7=0$
$\begin{array}{l}\therefore \frac{\mathrm{h}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{b}}=-\frac{\left({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ \Rightarrow \frac{\mathrm{h}-1}{1}=\frac{\mathrm{k}-5}{-2}=-\frac{(1-2(5)+7)}{1^2+(-2{)}^2}=\frac{2}{5}\\ \Rightarrow \frac{\mathrm{h}-1}{1}=\frac{2}{5},\frac{\mathrm{k}-5}{-2}=\frac{2}{5}\Rightarrow \mathrm{h}=\frac{2}{5}+1\\ \Rightarrow \mathrm{k}-5=\frac{-4}{5}\\ \mathrm{h}=\frac{7}{5}\mathrm{\&}\mathrm{k}=\frac{-4}{5}+5=\frac{21}{5}\\ \therefore \text{ Midpoint of the chord is }\mathrm{P}(\mathrm{h},\mathrm{k})=\left(\frac{7}{5},\frac{21}{5}\right)\end{array}$