Coordinate Geometry

$\sqrt{3^2 + 4^2} = \sqrt{25} = 5$. The distance formula is Pythagoras applied to the horizontal and vertical separations.

The signs flip out of the brackets, and the right side is $r^2$, so $r = \sqrt{16} = 4$.

Perpendicular gradients multiply to $-1$, so the required gradient is $-\dfrac{1}{3}$. Flip the fraction and change the sign.

$y - 3 = -2(x-2)$ gives $y = -2x + 4 + 3$. Check: at $x = 2$, $y = 3$ ✓

Here $2g = 8$ and $2f = -2$, so the centre is $(-4, 1)$, and $r = \sqrt{16 + 1 + 8} = 5$.

The midpoint is $(3,6)$ and the segment's gradient is 2, so the bisector has gradient $-\tfrac12$: $y - 6 = -\tfrac12(x-3)$, giving $y = -\tfrac12 x + 7.5$. Check: at $x = 3$, $y = 6$ ✓