stuck on a Cartesian question we have a circle $(x-1)^2+(y-2)^2=9$
Point $P=(5,2)$ lies outside the circle.
Solve the equation of the line which passes through $P$ and intersects the circle at two points whose mutual distance is $d=2$. Find the coordinates of the intersection points.
I sketched the circle and $P$ but could not find the relationship between these.
Please help me to get the intersection points.
The answer below is quite confused and is not correct. Could someone help me ?

HINT:
Let the equation of the line be $$\frac{y-2}{x-2}=m\implies y=mx+2-2m\ \ \  \ (1)$$ where $m$ is the gradient
Replacing the value of $y$ in  $$(x-1)^2+(y-2)^2=9$$
$$(x-1)^2+(mx-m)^2=9\iff x^2(1+m^2)-2x(1+m^2)+m^2-8=0\ \ \ \ (2)$$
If $(x_1,y_1);(x_2,y_2)$ are the points of intersection, we have $\displaystyle x_1+x_2=\frac{2(1+m^2)}{1+m^2}=2$ and $\displaystyle x_1x_2=\frac{m^2-8}{1+m^2}$
So, if $d$ is the distance between the two points,
$$d^2=(x_1-x_2)^2+(y_1-y_2)^2=(1+m^2)(x_1-x_2)^2=(1+m^2)\{x_1+x_2)^2-4x_1x_2\}$$
$$=(1+m^2)\left(2^2-4\frac{(m^2-8)}{1+m^2}\right)$$
Find $m$ from the given condition , then the two values of $x$ from $(2)$ and those of $y$s from $(1)$
