Question 815686
The second part gives you a circle, just using circle equation knowledge, being {{{(x-5)^2+(y-3)^2=25}}};  the first part, depends how clever you are.  You can use distance formula to develop an equation for it....


Three units from y-axis or from a general point (0,y).
Distance Formula:   {{{sqrt((x-0)^2+(y-y)^2)=3}}}, but not really necessary.
You know "3 units from the y-axis", so {{{x=-3}}} or {{{x=3}}}.  This can also be stated, {{{x^2=9}}}.


This plan is now to solve the circle for y, treating x as a constant.
Then, solve for y using separately x=-3 and for x=3.  I believe these will be two solutions *, two points.  To get started,
{{{(y-3)^2=-(x-5)^2+25}}}
{{{y-3=0+- sqrt(25-(x-5)^2)}}}
{{{highlight(y=3+- sqrt(25-(x-5)^2))}}}-------partway toward solution.
.
at which extent you can now use both values for x and compute the corresponding values for y.


*Note that this has ONE Real solution.  The choice of {{{x=-3}}} will not work because the square root will become {{{sqrt(25-(-3-5)^2)=sqrt(25-64)=sqrt(-39)}}}, which is Complex, not Real.  If you must have REAL number coordinates, then you can only choose {{{x=3}}}.