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

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  1. r < 2
  2. r = 2
  3. r > 2
  4. 2 < r < 8

Answer (Detailed Solution Below)

Option 4 : 2 < r < 8
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Detailed Solution

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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 = r.... (1)

And the equation of circle Q is,

x2 + y- 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 = 

For the intersection of circles,

D > |rP - rQ| & D < |rP + rQ| .... (3)

Here rP & rQ is the radius of both the circle,

We have,

rP = r

& rQ = 3

From equation (3),

5 > |r - 3|; r < 8

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

Hence, 2 < r < 8

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