The circle touches the co-ordinate axes and the line x = a is shown in the diagram.
\(\therefore \ centre\ is\ \left( \frac { a }{ 2 } ,\frac { a }{ 2 } \right) and\ r=\frac { a }{ 2 } \)
\(\therefore\) There may be two such circles, one lying

above X-axis and other below X-axis.
Equation of the circle lying above the X-axis is
\({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

Equation of the circle lying below the X-axis is \({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

Let (3, -4) and radius = Let C(3, -4) be the centre of the circle. If the line 5x + 12y - 12 = 0 touches the circle at P, then radius = perpendicular distance from p to the line 5x + 12y - 12 = 0.
\(\therefore r=\left| \frac { 5(3)+12(-4)-12 }{ \sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } } \right| =\frac { 45 }{ 13 } \)
\(\therefore\) Equation of the circle with centre 
(3,-4) and radius= \(\frac { 45 }{ 13 } \quad is\)
(x - 3)² + (y + 4)² = \({ \left( \frac { 45 }{ 13 } \right) }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+6x+9+{ y }^{ 2 }+8y+16=\frac { 2025 }{ 169 } \)
\(\Rightarrow \) 169 x² + 1014 x + 1521 + 169y² + 1352 y + 2704 = 2025
\(\Rightarrow \) 169 x² + 169 y² + 1014x + 1352y + 2200 = 0

Let P(x₁, y₁) be any point on the locus and let A(0, 2) B(0, -2) be the fixed points.
By the given condition, PA + PB =6
\(\Rightarrow \sqrt { { \left( { x }_{ 1 }-0 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 } } +\sqrt { { ({ x }_{ 1 }-0 })^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } =6\)
\(\Rightarrow \sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 } } =6-\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring both sides we get,
\({ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }-36-{ x }_{ 1 }^{ 2 }-{ y }_{ 1 }^{ 2 }-4-4{ y }_{ 1 }=-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= -8y₁ - 36 = -12 \(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= 2y₁ + 9 = 3\(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring again we get,
(2y₁+9)² = 9[\({ x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }-9{ x }_{ 1 }^{ 2 }-9{ y }_{ 1 }^{ 2 }-36-36{ y }_{ 1 }=0\)
\(\Rightarrow -9{ x }_{ 1 }^{ 2 }-5{ y }_{ 1 }^{ 2 }+45=0\)
\(\Rightarrow 9{ x }_{ 1 }^{ 2 }+5{ y }_{ 1 }^{ 2 }=45\)
\(\therefore \)  Locus of (x₁ , y₁) is 9x² + 5y² = 45

The given lines are
y = m₁x + C₁ \(\Rightarrow\) m₁ x - y + C₁ = 0
y = m₂x + C₂ \(\Rightarrow\) m₂x - y + C₂ = 0
y = m₃x + C₃ \(\Rightarrow\) m₃x - y + C₃ = 0.
The condition for these lines to be concurrent is \(\left| \begin{matrix} { m }_{ 1 } & -1 & { C }_{ 1 } \\ { m }_{ 2 } & -1 & { C }_{ 2 } \\ { m }_{ 3 } & -1 & { C }_{ 3 } \end{matrix} \right| =0\)
Interchanging C₂ and C₃ we get,
\(\left| \begin{matrix} { m }_{ 1 } & { C }_{ 1 } & 1 \\ { m }_{ 2 } & { C }_{ 2 } & 1 \\ { m }_{ 3 } & { C }_{ 3 } & 1 \end{matrix} \right| =0\)    [Taking (-1) common from C₃]
Expanding along C₁ we get
\({ m }_{ 1 }\left| \begin{matrix} { C }_{ 2 } & 1 \\ { C }_{ 3 } & 1 \end{matrix} \right| -{ m }_{ 2 }\begin{vmatrix} { C }_{ 1 } & 1 \\ { C }_{ 3 } & 1 \end{vmatrix}+{ m }_{ 3 }\begin{vmatrix} { C }_{ 1 } & 1 \\ { C }_{ 2 } & 1 \end{vmatrix}=0\)
\(\Rightarrow\) m₁(C₂ - C₃) - m₂(C₁ - C₃) + m₃(C₁ - C₂) = 0
\(\Rightarrow\) m₁(C₂ - C₃) + m₂(C₃ - C₁) + m₃(C₁ - C₂) = 0

The centre is the point of intersection of the diameters.
Solving 2x - 3y + 12 = 0 ....(1)  and
x + 4y - 5 = 0  ....(2)
(1)     \(\rightarrow\)    2x - 3y + 12 = 0
                           -      -      +
(2)\(\times\)2 \(\rightarrow\)   2x + 8y - 10 = 0
                   ___________________
                       -11y + 22 = 0
\(\Rightarrow \) -11y = -22 
\(\Rightarrow \) y = 2
Substituting y = 2 in (2) we get,
x + 4(2) - 5 = 0 
\(\Rightarrow \) x + 8 - 5 = 0 
\(\Rightarrow \) x + 3 = 0 
\(\Rightarrow \) x = -3.
\(\therefore \)  (-3, 2) is the center of the circle.
Also, given area = 154  \(\Rightarrow \) \(\pi\)r² = 154
\(\Rightarrow \frac { 22 }{ 7 } \times { r }^{ 2 }=5\)
\(\Rightarrow { r }^{ 2 }=\frac { 154\times 7 }{ 22 } =\frac { 14\times 7 }{ 2 } \)
\(\Rightarrow\) r² = 49  \(\Rightarrow\) r = 7.
\(\therefore \)  Equation of the circle is (x + 3)² + (y - 2)² = 49
\(\Rightarrow\)  x² + 6x + 9 y² - 4y + 4 = 49
\(\Rightarrow\)  x² + y² + 6x - 4y - 36 = 0.

\(\Rightarrow\) Let the co-ordinates of the centre of the required circle be C(a,0). Since it passes through P(2,3) = 25
CP = radius = 5
\(\Rightarrow \sqrt { ({ a-2) }^{ 2 }+{ (0-3) }^{ 2 } } =5\)
\(\Rightarrow\) (a- 2 )² + 9 = 25 \(\Rightarrow\) (a - 2)² = 16
\(\Rightarrow\) (a - 2)² = (\(\pm \)4)²

\(\Rightarrow\) a - 2 = \(\pm \)4
\(\Rightarrow\) a =   \(\pm \) 4 + 2
\(\Rightarrow\) a = 4 + 2 or -4 + 2
\(\Rightarrow\) a = 6 or -2
Thus the Co-ordinates of the centre are (6, 0) or (-2, 0)
Hence the equations of the required circle are
(x - 6)² + (y - 0)² = 5² \(\Rightarrow\) x² + 36 - 12x + y² = 25
\(\Rightarrow\) x² + y² - 12x + 11 = 0 (OR)
(x + 2)² + (y - 0)² = 5²
\(\Rightarrow\) x² + 4x +4 + y² = 25
\(\Rightarrow\)  x² + y² + 4x - 21 = 0.

Let p(x₁,y₁) be the point on the locus and A(-1, 0) B(0, 2) are the fixed points
Given pA = 3 pB
\(\Rightarrow\)  pA² = 9 pB²
\(\Rightarrow ({ x }_{ 1 }+1)^{ 2 }+({ y }_{ 1 }-0)^{ 2 }=9[({ x }_{ 1 }-0)^{ 2 }+({ y }_{ 1 }-2)^{ 2 }]\)
\(\Rightarrow { x }_{ 1 }^{ 2 }+2{ x }_{ 1 }+1+{ y }_{ 1 }^{ 2 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-4{ y }_{ 1 }+4]\)
\(\Rightarrow { x }_{ 1 }^{ 2 }+2{ x }_{ 1 }+1+{ y }_{ 1 }^{ 2 }=9{ x }_{ 1 }^{ 2 }+9{ y }_{ 1 }^{ 2 }-36{ y }_{ 1 }+36\)
\(\Rightarrow 8{ x }_{ 1 }^{ 2 }+8{ y }_{ 1 }^{ 2 }-{ 2x }_{ 1 }-36{ y }_{ 1 }+35=0\)
\(\therefore Locus\quad of\quad (x_{ 1 },{ y }_{ 1 })\quad is\) 8x² + 8y² - 2x - 36y + 35 = 0

The length of the tangent to the circle \({ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0\) from a point \(\left( { x }_{ 1 },{ y }_{ 1 } \right) \) is \(\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+2{ gx }_{ 1 }+2{ fy }_{ 1 }+c } \) 
Length of the tangent
\(\\ =\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+8{ x }_{ 1 }+4{ y }_{ 1 }+8 } \)
\(=\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+8(2)+4(3)+8 } \)
\( \left[ Here({ x }_{ 1 }{ y }_{ 1 })=(2,3) \right] \)
\(=\sqrt { 49 } \)
Length of the tangent = 7 units

The equation of the tangent at (x₁, y₁) to the given circle \({ x }^{ 2 }+{ y }^{ 2 }+3x-8y+17=0\) is
\({ xx }_{ 1 }+{ yy }_{ 1 }+3\times \cfrac { 1 }{ 2 } \left( x+{ x }_{ 1 } \right) -8\times \cfrac { 1 }{ 2 } \left( y+{ y }_{ 1 } \right) +17=0\) 
Here \(\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -2,5 \right) \) 
\(-2x+5y+\cfrac { 3 }{ 2 } \left( x-2 \right) -4(y+5)+17=0\)
\( -2x+5y+\cfrac { 3 }{ 2 } x-3-4y-20+17=0\)
\(-4x+10y+3x-6-8y-40+34=0\) 
\(x-2y+12=0\) is the required equation.

The circle \({ x }^{ 2 }+{ y }^{ 2 }+ax+by=0\) passing through (1, 2) and (1, 1)
\(\therefore \) We have 1 + 4 + a + 2b = 0 and  1 + 1 + a + b = 0
\(\Rightarrow a+2b=-5\) (1)
and a + b = –2  (2)
Solving (1) and (2),we get a = 1,b = -3