The line $y+2x=11$ is a tangent to a circle with a centre $P(1;-1)$ at the point $Q(x;y)$. Determine the equation of the radius $PQ$

Question

I already find the radius to be $10$ units. I did that by substituting $p$ values aka the centre into the equation of a circle

Answer 1

1. With your given $y=-2x+11$, you immediately have a slope of $(-\frac{1}{-2}x)=\frac{1}{2}x$ for its perpendicular through $Q$. The explicit form of this perpendicular is $y=\frac{1}{2}x+c$, with $c$ such that it crosses $P=(1,-1)$. We need to shift $y$ by $c=\frac{-3}{2}$, so

$y=\frac{1}{2}x-1.5$.

2. Equating the given tangent line and its now established perpendicular through $P$ yields the intersection point

$Q=(5,1)$.

3. As for the radius $r=PQ$, we know it satisfies $(x-1)^2+(y+1)^2=r^2$. Let us either insert $Q$ or establish that the vector difference between $P$ and $Q$ is $(4,2)$ - either way, an equation for radius r is

$r=\sqrt{(x_Q-1)^2+(y_Q+1)^2}$ $r=\sqrt{4^2+2^2}=\sqrt{20}$.