Two diameters are along

$$2x+3y+1=0$$ and $$3x-y-4=0$$

solving we get center $$(1,-1)$$

circumference $$ = 2\pi r = 10\pi $$

$$\therefore$$ $$r=5$$.

Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$

$$\pi {r^2} = 154 \Rightarrow r = 7$$

For center on solving equation

$$2x - 3y = 5\& 3x - 4y = 7$$

we get $$x = 1,\,y = - 1$$

$$\therefore$$ center $$=(1,-1)$$

Equation of circle,

$${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$$

$${x^2} + {y^2} - 2x + 2y = 47$$

$$\left| {{r_1} - {r_2}} \right| < {C_1}{C_2}$$ for intersection

$$ \Rightarrow r - 3 < 5 \Rightarrow r < 8\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $${r_1} + {r_2} > {C_1}{C_2},\,$$

$$r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)$$

From $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$2 < r < 8.$$

For any point $$P(x,y)$$ in the given circle,



we should have

$$OA \le OP \le OB$$

$$ \Rightarrow \left( {5 - 3} \right) \le \sqrt {{x^2} + {y^2}} \le 5 + 3$$

$$ \Rightarrow 4 \le {x^2} + {y^2} \le 64$$