Yes?????

Because your title literally states there are 5 points on each circle

gundamchamp wrote: Because your title literally states there are 5 points on each circle That's the problem statement: we want to have five points on each circle, five circles through each point. It doesn't mean that a construction is necessarily possible.

Here is an amusing fake disproof that the answer is No! (As far as I know, the answer is Yes but I have not looked for a construction.) Consider the bipartite graph $G$ with classes $C$ and $P$ in which there is an edge between a circle in $C$ and a point in $P$ if the point lies on the circle. The condition requires all vertices to be of degree $5$. By counting the number of edges emanating from $C$ and those from $P$ we now get $5|C| = 5|P|$, i.e. $k:= |C| = |P|$ and the edges are $5k$ in amount. Next, observe that $G$ does not contain a copy of $K_3$ (since it is bipartite), hence also does not contain $K_5$. Moreover, it is not possible that three distinct points lie simultaneously on three distinct circles, hence $G$ does not contain $K_{3,3}$. It follows by Kuratowski's theorem that $G$ is planar with no triangles and hence has at most $2(2k-2) = 4k-4$ edges. But $4k - 4 < 5k$, contradiction. (The fault is in the misuse of the statement of Kuratowski's theorem - subdivision issues.)

Marinchoo wrote: Several circles are drawn on the plane and all points of their intersection or touching are marked. Is it possible that each circle contains exactly five marked points and each point belongs to exactly five circles?

Consider a icosahedron and the 12 circles passing through 5 of its vertices. The stenografic projection of the exsphere with the 12 points and 12 circles has the desired properties.

I have a construction with circles and lines. Not sure that's allowed.

The official solution remarks the following: For any $k=2,3,4,5$ there exists a configuration of several circles and their common points, such that each circle passes through exactly $k$ points and each point belongs to exactly $k$ circles. It is not known does such configurations exist for $k > 5$. Below are the respective constructions for $k=2,3,4,5$ (note we will be using the fact that $k$ of the circles can be assumed to be lines, due to inversion): [asy][asy] size(300); pair O = (0,0); for(int i=1 ; i < 6 ; ++i){ draw(circumcircle(O,dir(18+72*i),dir(18+72*(i+2))),fuchsia); draw(extension(dir(18+72(i+1)),dir(18 + 72(i+2)),dir(18 + 72(i-1)),dir(18+72(i-2)))--extension(dir(18+72(i+2+1)),dir(18 + 72(i+2+2)),dir(18 + 72(i+2-1)),dir(18+72(i+2-2))),orange); draw(circle(O,abs(O-dir(90))),cyan); draw(circle(O,abs(O-extension(dir(18+72(1+1)),dir(18 + 72(1+2)),dir(18 + 72(1-1)),dir(18+72(1-2))))),green); } dot(O); for(int i=1 ; i < 6 ; ++i){ pair Ai = dir(18+72*i); dot(Ai); } for(int i = 1 ; i < 6 ; ++i){ pair Bi = extension(dir(18+72(i+1)),dir(18 + 72(i+2)),dir(18 + 72(i-1)),dir(18+72(i-2))); dot(Bi); } [/asy][/asy] \begin{align*} \text{Construction for } k=5 \end{align*} [asy][asy] size(300); pair O = (0,0); for(int i=1 ; i < 5 ; ++i){ draw(circumcircle(O,dir(90*(i+1)),dir(90*i)),fuchsia); draw(2*foot(O,dir(90*(i+1)),dir(90*i))-O--2*foot(O,dir(90*(i+1)),dir(90*(i+2)))-O,orange); } draw(circle(O,1),cyan); draw(circle(O,sqrt(2)),green); dot(O); for(int i=1 ; i<5 ; ++i){ pair Ai=dir(90*i); dot(Ai); pair Bi=2*foot(O,dir(90*(i+1)),dir(90*i))-O; dot(Bi); } [/asy][/asy] \begin{align*} \text{Construction for } k=4 \end{align*} [asy][asy] size(300); draw(dir(90)--dir(210)--dir(-30)--dir(90),orange); draw(unitcircle,green); dot(dir(90)^^dir(210)^^dir(-30)); [/asy][/asy] \begin{align*} \text{Construction for } k=3 \end{align*} [asy][asy] size(300); draw(unitcircle,cyan); draw(shift(dir(0))*unitcircle,green); dot(dir(60)^^dir(-60)); [/asy][/asy] \begin{align*} \text{Construction for } k=2 \end{align*}