Any circle passing through the points of intersection of the given line and circle has the equation $${x^2} + {y^2} - 9 + \lambda \left( {x + y - 1} \right) = 0.$$       Its centre $$ = \left( { - \frac{\lambda }{2},\, - \frac{\lambda }{2}} \right).$$
The circle is the smallest if $$\left( { - \frac{\lambda }{2},\, - \frac{\lambda }{2}} \right)$$   is on the chord $$x + y = 1.$$
$$\therefore \,\, - \frac{\lambda }{2} - \frac{\lambda }{2} = 1\,\,\,\,\, \Rightarrow \lambda = - 1$$
Putting this value for $$\lambda ,$$ the equation of the smallest circle is $${x^2} + {y^2} - 9 - \left( {x + y - 1} \right) = 0.$$

$$\eqalign{ & {\text{The centre}} = \left( {t,\,t} \right) \cr & {\text{The radius}}\,\, = 2\sqrt 2 = \left| {\frac{{t + t - 4}}{{\sqrt 2 }}} \right|.\,\,{\text{So, }}\left| {2t - 4} \right| = 4 \cr & \therefore \,\,2t - 4 = \pm 4,\,{\text{i}}{\text{.e}}{\text{.,}}\,t = 4,\,0.\,\,{\text{So, the centre}} = \left( {4,\,4} \right),\,\left( {0,\,0} \right) \cr & \therefore \,\,{\text{satisfying}}\,\,x + y > 4,\,{\text{the centre}} = \left( {4,\,4} \right). \cr} $$

The radical axis of the circles is $$x - y = 0.$$   So, $$PA = PB = 3.$$

The centre of two circles are $${C_1}\left( {2,\,2} \right)$$   and $${C_2}\left( {0,\,0} \right).$$   The radii of two circles are $${r_1} = 2\sqrt 2 $$   and $${r_2} = 4$$

The equation of the common chord of the circles $${x^2} + {y^2} - 4x - 4y = 0$$     and $${x^2} + {y^2} = 16$$   is $$x + y = 4$$   which meets the circle $${x^2} + {y^2} = 16$$   at points $$A\left( {4,\,0} \right)$$  and $$B\left( {0,\,4} \right).$$  Obviously $$OA \bot OB.$$   Hence, the common chord $$AB$$  makes a right angle at the centre of the circle $${x^2} + {y^2} = 16.$$   Where, $$O$$ is the origin and the centre $${C_2}$$ of the second circle.

As per question area of one sector $$= 3$$  area of another sector
$$ \Rightarrow $$  angle at centre by one sector $$ = 3 \times $$  angle at centre by another sector
Let one angle be $$\theta $$ then other $$ = 3\theta $$
Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$
$$\therefore $$  Angle between the diameters represented by combined equation $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0{\text{ is }}{45^ \circ }$$
$$\therefore $$  Using $$\tan \,\theta = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}}$$
we get, $$\tan \,{45^ \circ } = \frac{{2\sqrt {{{\left( {a + b} \right)}^2} - ab} }}{{a + b}}$$
$$\eqalign{ & \Rightarrow 1 = \frac{{2\sqrt {{a^2} + {b^2} + ab} }}{{a + b}} \cr & \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right) \cr & \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab \cr & \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0 \cr} $$

From the figure, the centers are
$$\eqalign{ & \left( {1 \pm 5\cos \,\theta ,\,1 \pm 5\sin \,\theta } \right),{\text{ where}} \cr & \tan \,\theta = - \frac{1}{{ - \frac{3}{4}}} = \frac{4}{3} \cr & {\text{So, }}\cos \,\theta = \frac{3}{5},\,\,\sin \,\theta = \frac{4}{5} \cr} $$

Length of chord
$$\eqalign{ & = 2{\left\{ {{{\left( {{\text{radius}}} \right)}^2} - {{\left( {{\text{length of }} \bot \,{\text{from centre of chord}}} \right)}^2}} \right\}^{\frac{1}{2}}} \cr & = 2{\left\{ {{r^2} - {{\left( {\frac{{ - 1}}{{\sqrt {\left( {\frac{1}{{{a^2}}}} \right) + \left( {\frac{1}{{{b^2}}}} \right)} }}} \right)}^2}} \right\}^{\frac{1}{2}}} \cr & = 2\sqrt {\frac{{{r^2}\left( {{a^2} + {b^2}} \right) - {a^2}{b^2}}}{{{a^2} + {b^2}}}} \cr} $$

The length of the perpendicular from the centre (0, 0) to the line $$ = \frac{2}{{\sqrt {1 + {m^2}} }}$$
The radius of the circle $$= 1.$$  For the line to cut the circle at distinct or coincident points, $$\frac{2}{{\sqrt {1 + {m^2}} }} \leqslant 1{\text{ or }}1 + {m^2} \geqslant 4{\text{ or }}{m^2} \geqslant 3.$$

Let centre of the circle be $$\left( {1,\,h} \right)$$  [ $$\because $$ circle touches $$x$$-axis at $$\left( {1,\,0} \right)$$  ]

Let the circle passes through the point $$B\left( {2,\,3} \right)$$
$$\eqalign{ & \therefore CA = CB\,\,\,\,\,\,\,\left( {{\text{radius}}} \right) \cr & \Rightarrow C{A^2} = C{B^2} \cr & \Rightarrow {\left( {1 - 1} \right)^2} + {\left( {h - 0} \right)^2} = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2} \cr & \Rightarrow {h^2} = 1 + {h^2} + 9 - 6h \cr & \Rightarrow h = \frac{{10}}{6} = \frac{5}{3} \cr} $$
Thus, diameter is $$2h = \frac{{10}}{3}$$

Since circle touches $$x$$-axis at $$\left( {3,\,0} \right)$$
$$\therefore $$ The equation of circle be $${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$

As it passes through $$\left( {1,\, - 2} \right)$$
$$\therefore $$ Put $$x=1,\,\,y=-2$$
$$\eqalign{ & \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0 \cr & \Rightarrow \lambda = 4 \cr} $$
$$\therefore $$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$
Now, from the options $$\left( {5,\, - 2} \right)$$  satisfies equation of circle.