Question

Let $x-2y+7=0$ be a chord of the circle $x^2+y^2-2x-10y+1=0$. If the midpoint of the chord is $P(\alpha ,\beta )$, then the value of $5|\alpha -\beta |$ is

Answer 1

Given chord is $x-2y+7=0$
$P=(\alpha ,\beta )$
Now, $\alpha -2\beta +7=0\cdots \left(1\right)$
$CP$ is perpendicular to chord, so
$\frac{\beta -5}{\alpha -1}=-2$
Using equation $\left(1\right)$, we get
$\beta -5=-2(2\beta -8)\Rightarrow \beta =\frac{21}{5}\Rightarrow \alpha =\frac{7}{5}\therefore 5|\alpha -\beta |=14$


Alternate method :
Given circle is $x^2+y^2-2x-10y+1=0$
Centre and radius is $C=(1,5),r=5$
$P$ is foot of perpendicular drawn from $(1,5)$ to $x-2y+7=0$, so
$\frac{\alpha -1}{1}=\frac{\beta -5}{-2}=-\left(\frac{1-10+7}{1^2+2^2}\right)\Rightarrow \frac{\alpha -1}{1}=\frac{\beta -5}{-2}=\frac{2}{5}\Rightarrow \alpha =\frac{7}{5},\beta =\frac{21}{5}\therefore 5|\alpha -\beta |=14$