Which of the circle the line px+qy+r=0 will intersect? If p, q and r are in arithmetic progression,
then the line 
px + qy + r = 0

necessarily intersects which of the following circles? 
$$x^2 + y^2 + 4x – 4y + 7 = 0$$
                         or
$$ x^2 + y^2 – 6x + 6y + 13 = 0 $$
I tried by assigning 


p=a
q= a+d
r =a+2d


where a = first term and d =common-difference.
I put these values in 


px + qy = r


And put this value of x in terms of y in the circle equations.
But then it get complicated.
And i got stuck.
Is there is any other easier way it can be solved ?
Thanks in advance.
 A: First, prove that regardless of the values of p, q and r, the line must pass through the point (1, -2).  Next, find out which of the two circles has this point on the inside.  You can then argue that the line must pass through this circle.
Post again if you need help with either of the two main steps, and I'll add more detail to this answer.
Edit
If p, q, r are in A.P., then $r=2q-p$.  So write $px+qy+r=0$ as $p(x-1)+q(y+2)=0$.  But $(1,-2)$ is always a solution to this, so this point must be on the line, regardless of $p$ and $q$.
A: Rewrite equations of the circles into following forms :
$C_1 : (x+2)^2+(y-2)^2=1$
$\Rightarrow (x_O,y_O)=(-2,2) ~\text{and}~ R_1=1 $
$C_2 : (x-3)^2+(y+3)^2=5 $
$\Rightarrow (x_O,y_O)=(3,-3) ~\text{and}~ R_2=\sqrt 5 $
Next , define distance of the center of circle from the line as :
$$d=\frac{|px_O+qy_O+r|}{\sqrt{p^2+q^2}}$$
Now , consider Intersection criterion :
$\begin{cases}
 \text{the line intersect a circle}, & \text{if }~ d<R\\
 \text{the line is tangent of circle}, & \text{if }~d=R\\
\text{line and circle have no common points}, & \text{if }~d>R
\end{cases}$
A: $x^2 + y^2 + 4x – 4y + 7 = 0$ can be written as $(x+2)^2+(y-2)^2=1^2$ and
$x^2 + y^2 – 6x + 6y + 13 = 0$ as $(x-3)^2+(y+3)^3 = {\sqrt{5}}^2$
The first circle has center $(-2, 2)$, radius $1$ and the second with center at $(3,-3)$  and radius $\sqrt{5}$.
These two circles do not intersect (why?) 
Also your equation of line that has coefficients in A.P as David suggested can be written as $p(x-1)+q(y+2)=0$, which always passes through $(1,-2)$ and among the two circles only the second circle passes through $(1,-2)$ 
Work on showing this, and also showing that it does not intersect the first circle.
