Question 1146372
The radical axis of two circles can be defined as the locus (set) of the points from where tangents to the two circles would have equal lengths.
It can also be defined as the locus of the points whose power relative to both circles is the same.
Power of a point relative to a circle is {{{d^2-r^2}}} , where
{{{d=distance}}}{{{to}}}{{{the}}}{{{circle}}}{{{center}}} , and
{{{r=radius}}}{{{of}}}{{{the}}}{{{circle}}} .
 
The definitions are equivalent, because in the right triangles formed by tangent, the radius, and the distance {{{d}}} .
The line segment of length {{{d}}} between the point and the circle center is the hypotenuse and the power of the point is the square of the length of the tangent segment,  
{{{drawing(300,300,-5.5,6.5,-2,10,
circle(-3,2,0.1),circle(3,2,0.1),
red(circle(-3,2,2)),green(circle(3,2,3)),
blue(triangle(-3,2,0,2,3,2)),blue(triangle(-0.4167,-2,-0.4167,3,-0.4167,12)),
triangle(-4,3.732,-0.4167,5.8,-1,5.464),triangle(-3,2,-4,3.732,-3.5,2.866),
triangle(-3,2,3,2,-0.4167,5.8),triangle(3,2,3.9083,4.8592,-0.4167,5.8),
locate(-3.3,3.2,red(r)),locate(3.2,4.2,green(r)),
locate(-1.7,4.1,red(d)),locate(1.3,3.6,green(d))
)}}}
 
UNDERSTANDING THE PROBLEM:
The equations of the circles are written in the general form.
They could be written in a form that allowed us to identify the center and radius of each circle.
For the first circle, that would be {{{(x-2)^2+(y-3)^2-(3+2^2+3^2)=0}}} <--> {{{(x-2)^2+(y-3)^2-16=0}}} , showing the center to be C(2,3) , and the radius to be {{{sqrt(16)=4}}} .
For the second circle, that would be {{{(x-6)^2+(y-7)^2-(6^2+7^2-65)=0}}} <--> {{{(x-6)^2+(y-7)^2-20=0}}} , showing the center to be C'(6,7) , and the radius to be {{{sqrt(20)}}} .
The equations in this form show a sum of squares that is the square of the distance from a point P(x,y) to the center of the circle, minus the square of the radius.
That is the power of point P(x,y) relative to the circle.
For points of a circle it is zero, their power relative to the circle is zero,
but for any point P(x,y) the power of P relative to the first circle can be written as
{{{(x-2)^2+(y-3)^2-16}}} , or as its equivalent form, {{{x^2+y^2-4x-6y-3}}} .
So, for all the points P(x,y) whose powers relative to both circles are the same,
{{{x^2+y^2-4x-6y-3=x^2+y^2-12x-14y+65}}} , and from there we get the equation of the radical axis of the two circles.  
{{{cross(x^2+y^2)-4x-6y-3=cross(x^2+y^2)-12x-14y+65}}} --> {{{-4x+12x-6y+14y=65+3}}} --> {{{8x+8y=68}}} --> {{{x+y=68/8}}} --> {{{highlight(y=8.5-x)}}}
The circles, with their centers, the line segment connecting those centers, and the radical axis are shown below.
 
{{{drawing(300,300,-3,11,-2,13,grid(1),
red(circle(2,3,4)),green(circle(6,7,4.472)),
red(circle(2,3,0.2)),green(circle(6,7,0.2)),
blue(line(-3,11.5,11,-2.5)),blue(line(2,3,6,7)) 
)}}}