I find some answers to find three or more pythagorean triples with the same value of $c$, but how can i find some other pairs of primitive pythagorean triples with the same $c$? For example, $33^2+56^2 = 16^2 +63^2 = 65^2$, $36^2+77^2 = 13^2 + 84^2 = 85^2$.
Each primitive triple is $(a,b,c)=(u^2-v^2,2uv,u^2+v^2)$ where $u$ and $v$ satisfy some extra conditions. As $c$ must be odd, we need to find odd numbers which are sums of two squares in more than one way. There is a big theory on sums of squares, and it is known that if $c=pq$ where $p$ and $q$ are distinct primes congruent to $1$ modulo $4$ then $c$ can be written as a sum of two squares in two essentially different ways. So we can have $c=65$, $85$, $145$ etc.
So $145=1^2+12^2$ and $145=8^2+9^2$ so we get Pythagorean triples $143^2+24^2=145^2$ and $17^2+144^2=145^2$.