Question 992326
{{{(x-3)^2 + (y-4)^2 = 25}}}
{{{(x-1)^2 + (y-5/2)^2 = 225/4}}}<pre>
Subtract the two equations to get the equation of the line through
their point(s) of intersection if they have any points in common. 

{{{(x - 3)^2 - (x-1)^2  + (y-4)^2 - (y-5/2)^2 =25-225/4}}}

Factor first two and last two terms on the left as the
difference of squares:

{{{((x-3)^""-(x-1)^"")((x-3)^""+(x-1)^"")+((y-4)^""-(y-5/2)^"")((y-4)^""+(y-5/2)^"")=100/4-225/4}}}

{{{(x-3-x+1)(x-3+x-1)+(y-4-y+5/2)(y-4+y-5/2)=-125/4}}}

{{{(-2)(2x-4)+(-4+5/2)(2y-4-5/2)=-125/4}}}

{{{(-2)(2x-4)+(-8/2+5/2)(2y-8/2-5/2)=-125/4}}}

{{{-4x+8+(-3/2)(2y-13/2)=-125/4}}}

{{{-4x+8-3y+39/4=-125/4}}}

Multiply through by 4

{{{-16x+32-12y+39=-125}}}

{{{-16x-12y+71=-125}}}

{{{-16x-12y+71=-125}}}

{{{-16x-12y=-196}}}

Divide through by -4

{{{4x+3y=49}}}

{{{3y=-4x+49}}}

{{{y=expr(-4/3)x+49/3}}}

This is the equation of the line through their
point(s) of intersection if the intersect. 

Substitute in

{{{(x-3)^2 + (y-4)^2 = 25}}}

{{{x^2-6x+9+ ((expr(-4/3)x+49/3)^""-4)^2 = 25}}}

{{{x^2-6x+9+ (expr(-4/3)x+49/3-4)^2 = 25}}}

{{{x^2-6x+9+ (expr(-4/3)x+49/3-12/3)^2 = 25}}}

{{{x^2-6x+9+ (expr(-4/3)x+37/3)^2 = 25}}}

{{{x^2-6x+9+ (expr(1/3)(-4x+37)^"")^2 = 25}}}

{{{x^2-6x+9+ expr(1/9)(-4x+37)^2 = 25}}}

{{{x^2-6x+9+ expr(1/9)(16x^2-296x+1369) = 25}}}

Multiply through by 9

{{{9x^2-54x+81+(16x^2-296x+1369) = 225}}}

{{{9x^2-54x+81+16x^2-296x+1369 = 225}}}

{{{25x^2-350x+1450 = 225}}}

Divide through by 25

{{{x^2-14x+58 = 9}}}

{{{x^2-14x+49 = 0}}}

{{{(x-7)^2=0}}}

That give the double solution x=7.  The double solution
means that the circles intersect at one point, where x=7.

Substituting in

{{{4x+3y=49}}}
{{{4(7)+3y=49}}}
{{{28+3y=49}}}
{{{3y=21}}}
{{{y=7}}}

Therefore they intersect at the one point (7,7)

{{{drawing(400,400,-9,11,-7,13,
circle(7,7,0.15),circle(7,7,0.13),circle(7,7,0.11),circle(7,7,0.09),circle(7,7,0.07),circle(7,7,0.05),circle(7,7,0.03),circle(7,7,0.01),
locate(7.5,7.5,"(7,7)"),

graph(400,400,-9,11,-7,13),  circle(3,4,5), circle(1,5/2,15/2) )}}}

Edwin</pre>