From the line equation, we can deduce
[tex]x=-4y[/tex]
This implies that any point on the line can be written as
[tex](-4y,y),\quad y \in \mathbb{R}[/tex]
So, given any two points
[tex]A=(-4y_1,y_1),\quad B=(-4y_2,y_2)[/tex]
their distance will be
[tex]AB=\sqrt{(-4y_1+4y_2)^2+(y_1-y_2)^2}=\sqrt{17(y_1-y_2)^2}=\sqrt{17}|y_1-y_2|[/tex]
So, we can compute the distance between two points on the line just by knowing their y coordinates!
Now, the equation of a circle with center [tex](-1,2)[/tex] and radius 5 is
[tex](x+1)^2+(y-2)^2=5^2[/tex]
which we can reform as
[tex]x^2 + y^2 + 2 x - 4 y - 20 = 0[/tex]
The points of intersection between the circle and the line are given by the system between the two equations:
[tex]\begin{cases}x^2 + y^2 + 2 x - 4 y - 20 = 0\\x+4y=0\end{cases}[/tex]
From the second equation we can deduce [tex]x=-4y[/tex]. Plugging this value in the first equation yields
[tex](-4y)^2 + y^2 + 2 (-4y) - 4 y - 20 = 0 \iff 17y^2-12y-20=0[/tex]
Solving this equation for [tex]y[/tex] yields
[tex]y=\dfrac{12\pm\sqrt{144+1360}}{34}=\dfrac{12\pm\sqrt{1504}}{34}=\dfrac{12\pm\sqrt{16\cdot 94}}{34}=\dfrac{12\pm4\sqrt{94}}{34}=\dfrac{6\pm2\sqrt{94}}{17}[/tex]
Now we have the two [tex]y[/tex] values
[tex]y_1=\dfrac{6+2\sqrt{94}}{17},\quad y_2=\dfrac{6-2\sqrt{94}}{17}[/tex]
Which implies that the distance length of AB is
[tex]AB=\sqrt{17}|y_1-y_2|=\sqrt{17}\cdot\dfrac{4\sqrt{94}}{17}=4\sqrt{\dfrac{94}{17}}[/tex]
