Finding Points of Intersection of 2 circles Sorry for the really basic question but this has been nagging me for a while.
I have two circles:
$ (x-2)^2 + (y-3)^2 = 9$ (--eq1) and
$ (x-1)^2 + (y+1)^2 = 16$ (-- eq2)
Now, I am aware of the general method for solving this and getting the 2 solutions. 
When I rewrite the equations and subtract:
$ (x-2)^2 + (y-3)^2 = 9$
$ - ((x-1)^2 + (y+1)^2 = 16$) 

I get: $ x+2y = 9 $. This is a straight line passing through the 2 solutions. This makes sense. 
Now if I do:
eq1*16
eq2*9 
and equate the LHS I get: $ 7x^2 + 7y^2 - 46x -78y +190 = 0 $ 
When I plot this it's a (red) circle which doesn't pass through either solution point. Why doesn't this work? I have a vague intuition of what's going on but shouldn't the 2 solution points still satisfy the equations? I'm not looking for a vague answer for why this is wrong or what the correct solution is. I'd like to know at a deeper (maybe geometric) level for what's going on. Thanks in advance. 
I have attached the plots below:
Plots
Edit: $ 7x^2 + 7y^2 - 46x -78y +190 = 0 $ is wrong. Should be $ 7x^2 + 7y^2 - 46x -114y +190 = 0 $. My apologies. Been a long day. Circles intersect. Math makes sense.
Final Edit: Got the solution. Thank you for your responses everyone!:)
 A: When I equate the two LHS of the equations, I get$$7x^2-7y^2-46x-114y+190=0$$
which is not what you wrote.
A: Let's rewrite both ciurcle equations so the RHS is zero: 
$$
(x-2)^2 + (y-3)^2 - 9 = 0
$$
for instance. This has the form $f(x, y) = 0$; let's say the other is $g(x, y) = 0$. When you subtract one from the other, you get
$$
f(x, y) - g(x, y) = 0
$$
Now suppose that $(a, b)$ is a point on the first curve. Then you know that $f(a, b) = 0$; if it's also on the second, then $g(a, b) = 0$. That means that 
$$
f(a, b) - g(a, b) = 0,
$$
hence it lies on the third "curve" (which happens to be a line). So that certainly explains the first computation you did: you created the polynomial 
$f - g$, which happened to be linear, and contained the two intersection points. But suppose that instead of subtracting, you'd written down 
$$
pf(x, y) + q g(x, y) = 0
$$
where $p$ and $q$ are any two numbers. Then exactly the same analysis would get the same result: any such "linear combination" of the two defining polynomials gets you a new polynomial whose solutions contain any point $(a, b)$ that's a solution of both $f = 0$ and $g = 0$. In short:

If $f$ and $g$ are polynomials defining (via $f = 0$ and $g = 0$) the
  curves $C_f$ and $C_g$, and $X \in C_f \cap C_g$, then $X$ also lies
  on the curve defined by $pf + qg = 0$ for any real numbers $p$ and
  $q$.

In fact, the word "polynomial" is unnecessary in the theorem above -- they could just be functions. 
What you've discovered is that the various curves defined by $pf + qg = 0$ form a 
"pencil of curves", all of which contain the intersection points. If you're willing to restrict to the case $q \ne 0$, you can divide through by $q$ to say that this "pencil of curves" is the same one defined by $\frac{p}{q} f + g = 0$, or, more simply, $\alpha f + g = 0$, as $\alpha$ ranges over all real numbers. You might want to take your curve plot and add to it the curves for $\alpha = -2, -1, 0, 1, 2, $ and maybe a few more values (or use a Desmos "slider"!) to see what this "pencil of curves" looks like.
What you've done is snuck up on the very beginnings of some of the ideas of algebraic geometry --- nice work!
A: If $p.f(x,y)+q.g(x,y)=0$ (eq.0) ,then it's not necessary that $f(x,y)=g(x,y)=0$. I Think your curve is wrong , your circle should pass through your solution. Let take a simple example of straight lines since they are circles of infinite radius $x+y=1$  &  $2.x+y=2$ . we get multiplying eq. 1 by 2 and eq. 1 by 1 , $y=0$ and our solution is (1,0) which passes through $y=0$.
Also in method you use to solve it , you get a straight line which have infinite points on it not your solution , then you intersects it with one of your circles, to get your solution (Note: since one of f or g is zero then  eq.0 becomes zero only when second equation is also zero,which is required condition)
If you intersects your new circle one of circles you will get your desired solution.
A: Let:
$$ f=(x-2)^2 + (y-3)^2 = 9 \tag1$$
$$ g=(x-1)^2 + (y+1)^2 = 16 \tag2$$
$f-g=0$ where the circles intersect.
$$f-g=\left(x-2\right)^2\:+\:\left(y-3\right)^2-\left(\left(x-1\right)^2\:+\:\left(y+1\right)^2\right)+\left(-9+16\right)=0$$
Simplifying (not easy!): $$18-2x-8y=0$$ 
$$x=9-4y \tag3$$
Using either (1) or (2), say we use (2) with this value of $x$, to get:
$$(9-4y-1)^2+(y+1)^2-16=0$$
This has two solutions in $y$:
$$y=2.48904, y=1.15801$$
Using equation (3), with these values for $y$ we get 2 equations to solve for $x$:
$$18-2x-8(2.48904), 18-2x-8(1.15801)$$
Solving the above we get the 2 points of intersection as:
$$P1:(-0.95616
,2.48904), P2:(4.36796,1.15801)$$
Image for the problem and solution.
