Question 885215
For the first circle, the given equation is {{{x^2+y^2-3x-4y-1=0}}} .
We can transform it:
{{{x^2+y^2-3x-4y=1}}}
{{{x^2-3x+y^2-4y=1}}}
{{{x^2-3x+9/4+y^2-4y+4=1+9/4+4}}}
{{{(x-3/2)^2+(y-2)^2=1+9/4+4}}}
 
{{{x=3/2}}} and {{{y=2}}} are the coordinates of the center of the circle,
and {{{(x-3/2)^2+(y-2)^2}}} is the square of the distance from a point (x,y) to that center.
{{{1+9/4+4}}} is the square of the radius of the circle, but I do not need to calculate it.
 
The distance from the center of that first circle (with {{{x=3/2}}} and {{{y=2}}} ) 
to point (5,2) (also with {{{y=2}}} ) is {{{abs(5-3/2)=abs(7/2)=7/2}}} .
 
A circle centered at (5,2), with radius {{{R}}} has the equation
{{{(x-5)^2+(y-2)^2=R^2}}}
If the circle passes through a point with {{{x=3/2}}} and {{{y=2}}} ,
which is at a distance {{{7/2}}} from (5,2), then {{{R=7/2}}} , and
{{{(x-5)^2+(y-2)^2=(7/2)^2}}} is the equation of the second circle.
It can also be written as
{{{(x-5)^2+(y-2)^2=49/4}}} , or further manipulated into whatever equivalent equation you desire:
{{{x^2-10x+25+y^2-4y+4=49/4}}}-->{{{x^2+y^2-10x-4y+25+4-49/4=0}}}-->{{{x^2+y^2-10x-4y+67/4=0}}}-->{{{4x^2+4y^2-40x-16y+67=0}}}