If two circles (x - 1)² + (y - 3)² = r² and x² + y² - 8x + 2y + 8 = 0 intersect in two distinct points, then -

Concept:

The equation of the circle with centre at (h, k) and radius equal to a, is

(x - h)2 + (y - k)2 = a2

Calculation:

Consider the equation of circle P is,

(x - 1)2 + (y - 3)2 = r2 .... (1)

And the equation of circle Q is,

x2 + y2 - 8x + 2y + 8 = 0

It can be written as,

(x − 4)2 +(y + 1)2 = 9 = 3² .... (2)

From equation (1) & (2),

Coordinates of the center of both circles P & Q can be written as,
P ⇒ (1,3)
Q ⇒ (4,−1)

Hence, the distance between both circles (D) is given by,

D = \(\sqrt{(1-4)^2+(3-(-1))^2}=5\)

For the intersection of circles,

D > |r_P - r_Q| & D < |r_P + r_Q| .... (3)

Here r_P & r_Q is the radius of both the circle,

We have,

r_P = r

& r_Q = 3

From equation (3),

5 > |r - 3|; r < 8

& 5 < |r + 3|; r > 2

Hence, 2 < r < 8