Slope of line joining centres of circles = $${4 \over 3} = \tan \theta $$


$$ \Rightarrow \cos \theta = {3 \over 5},\sin \theta = {4 \over 5}$$

Now using parametric form

$${{x - 1} \over {\cos \theta }} = {{y - 2} \over {\sin \theta }} = \pm \,5$$

(x, y) = (1 + 5cos$$\theta$$, 2 + 5sin$$\theta$$)

($$\alpha$$, $$\beta$$) = (4, 6)

(x, y) = ($$\gamma$$, $$\delta$$) = (1 $$-$$ 5cos$$\theta$$, 2 $$-$$ 5sin$$\theta$$)

($$\gamma$$, s) = ($$-$$2, $$-$$2)

$$\Rightarrow$$ |($$\alpha$$ + $$\beta$$) ($$\gamma$$ + $$\delta$$)| = | 10x $$-$$ 4 | = 40

$$r = \sqrt {{{{p^2}} \over 4} + {{{{(1 - p)}^2}} \over 4} - 5} = {{\sqrt {2{p^2} - 2p - 19} } \over 2}$$

Since, $$r \in (0,5]$$

So, $$0 < 2{p^2} - 2p - 19 \le 100$$

$$ \Rightarrow p \in \left[ {{{1 - \sqrt {239} } \over 2},{{1 - \sqrt {39} } \over 2}} \right) \cup \left( {{{1 + \sqrt {39} } \over 2},{{1 + \sqrt {239} } \over 2}} \right]$$

so, number of integral values of p² is 61.

Let point P(x, y)

A(0, 0), B(1, 0), C(0, 1), D(1, 1)

(PA)² + (PB)² + (PC)² + (PD)² = 18

$${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$$ = 18

$$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$$

$$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$$

$$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $$

$$ \Rightarrow {d^2} = 16$$

$${S_1}:{(x - 5)^2} + {(y - 5)^2} = 9$$

Centre (5, 5), r₁ = 3

$${S_2}:{(x - 12)^2} + {(y - 5)^2} = 9$$

Centre (12, 5), r₂ = 3



So (P₁P₂)_min = 1