Common Chord of Two Circles

The circle $x^2+y^2=4$ cuts the circle $x^2+y^2+2x+3y-5=0$ in $A \& B$. Then the centre of the circle on $AB$ as a diameter is

The circles $x^2+y^2+2ax-c^2=0$ and $x^2+y^2+2bx-c^2=0$ intersect at $A$ and $B$. A line through $A$ meets one circle at $P$ and $a$a parallel line through $B$ meets the other circle at $Q$. Show that the locus of the mid point of $PQ$ is a circle.

If the circle $x^2+y^2+4x+22y+c=0$ bisects the circumference of the circle $x^2+y^2-2x+8y-d=0$, then $c+d$ is equal to:

Prove that the length of the common chord of the two circles $x^2+y^2=a^2$ and ${\left(x-c\right)}^2+y^2=b^2$ is $\frac{1}{c}\sqrt{\left(a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)\left(-a+b+c\right)}$, where $a,b,c>0$.

Show that the common chord of the circles $x^2+y^2-6x-4y+9=0$ and $x^2+y^2-8x-6y+23=0$ is the diameter of the second circle and also find its length.