<s>[INST] How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any?
 [/INST] Example 1: Find the points of intersection of the circles given by their equations as follows:
$(x - 2)^2 + (y - 3)^2 = 9$
$(x - 1)^2 + (y + 1)^2 = 16$
Solution to Example 1:
We first expand the two equations as follows: 
$x^2 - 4x + 4 + y^2 - 6y + 9 = 9 $
$x^2 - 2x + 1 + y^2 + 2y + 1 = 16 $
Multiply all terms in the first equation by -1 to abtain an equivalent equation and keep the second equation unchanged 
$-x^2 + 4x - 4 - y^2 + 6y - 9 = -9 $
$x^2 - 2x + 1 + y^2 + 2y + 1 = 16 $
We now add the same sides of the two equations to obtain a linear equation 
$2x - 3 + 8y - 8 = 7 $
Which may written as 
$x + 4y = 9 \textbf{ or } x = 9 - 4y $
We now substitute $x$ by $9 - 4y$ in the first equation to obatin 
$(9 - 4y)^2 - 4(9 - 4y) + 4 + y^2 - 6y + 9 = 9 $
Which may be written as 
$17y^2 -62y + 49 = 0 $
Solve the quadratic equation for y to obtain two solutions 
$y = \frac{(31 + 8\sqrt{2})}{17} \approx 2.49 $
and $ y =\frac{31 - 8\sqrt{2}}{17} \approx 1.16 $
We now substitute the values of y already obtained into the equation $x = 9 - 4y $ to obtain the values for x as follows 
$x = \frac{29 + 32\sqrt{2}}{ 17} \approx - 0.96 $
and $x = \frac{29 - 32\sqrt{2})}{17} \approx 4.37 $
The two points of intersection of the two circles are given by 
$(- 0.96 , 2.49)$ and $(4.37 , 1.16)$
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