How to find the radius if two circles intersect in two distinct points? Question- if two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8x+2y+8=0$  intersect in two distinct points , then find the range in which r exists 
I have these two circles
$(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8x+2y+8=0$ .
Now, I want to find out the range of r .
I have found out center $(4,-1)$ and radius as 3 of the second circle .
My book has mentioned a condition for circles to intersect at two places , we have.
$r_1-r_2<c_1c_2<r_1+r_2$
I don't understand what is this ?
My book says the answer is $2<r<8$
Please explain me the answer. Thank you !
 A: So you have a circle of center $(1,3)$ and radius $r$ and another circle of center $(4,-1)$ and radius $3$. For them to intersect in two places, two things have to happen: the interiors have to overlap (so $r$ cannot be too small) and the first circle cannot completely surround the second circle (so $r$ cannot be too big).
We can simplify the problem by shifting and rotating coordinates so that the first circle is centered at the origin and the second circle is centered on the positive $x$ axis. Then the second circle is now centered at $x=\sqrt{(1-4)^2+(-1-3)^2}=\sqrt{9+16}=5$ and $y=0$.
Having done that, notice that the point on the second circle which is closest to the center of the first circle is $(2,0)$ (the leftmost point). If we want the insides to overlap, then this point will have to be inside the first circle, so you will need $r>2$.
Now try to find a condition so that the first circle does not completely surround the second circle. Hint: what is the furthest point on the second circle from the center of the first circle?
A: For 2 circle intersecting each other as 2 points, the distance between centres $C_1$ and $C_2$ must be shorter than the distance when 2 circle are in contact (touching) for 1 point only, which is the case $C_1C_2=r_1+r_2$ as the centres form a straight line with the only intersecting point. So we have :
$$C_1C_2<r_1+r_2$$
While the minimum distance between centres exists when the smaller circle touches the larger one internally. So the minimum distance $C_1C_2\ge{r_1}-r_2$. However as the 2 circle intersects each other at 2 points, the case of touching  ($C_1C_2=r_1-r_2$) should be rejected. So 
$$C_1C_2>{r_1}-r_2$$
To conclude,
$$r_1-r_2<c_1c_2<r_1+r_2$$
A: The circles intersect at two distinct points,therefore
$|r_1-r_2|<$distance between centers of the circles$<|r_1+r_2|$
$$
|r-3|<\sqrt{(4-1)^2+(-1-3)^2}<|r+3|
$$
$$
|r-3|<5 \cap |r+3|>5
$$
You can solve after that.
