How long is the chord?

$\begin{aligned} x^2 + y^2 - 6x - 8y & = 0 & \small \color{#3D99F6} \text{Since }y = 2x+3 \\ x^2 + 4x^2+12x+9 - 6x - 16x-24 & = 0 \\ 5x^2 - 10x - 15 & = 0 \\ x^2 - 2x - 3 & = 0 \\ (x+1)(x-3) & = 0 \end{aligned}$

Since $y=2x+3$ , the end points of the chord are $(-1, 1)$ and $(3,9)$ and its length is $\sqrt{(3+1)^2+(9-1)^2} = \boxed{4\sqrt 5}$ .

Knowing that: $\\$ - The chord length $|AB| = 2\sqrt {r^2 - d^2}$ [ $d$ is the chord distace] $\\$ - $d = \frac {|Ax_0 + Bx_0 + C|}{\sqrt {A^2 + B^2}}$ $\\$

$x^2 + y^2 - 6x - 8y = 0 ==> (x-3)^2 + (y-4)^2 = 25$ $\\$ The circle with center $(3, 4)$ has chord distance $d = \frac {|2\times 3 - 4 + 3|}{\sqrt 5} = \sqrt 5$ $\\$ Also since $r = 5$ , $2\sqrt {r^2 - d^2} = 2\sqrt {25-5} = 4\sqrt 5$