\(\Rightarrow\) 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.

Given equation of pair of lines is
12x² + 7xy + ky² + 13x - y + 3 = 0
2h = 7   2g = 13   2f = -1
\(\Rightarrow a=12,\quad h=\frac { 7 }{ 2 } ,b=k,\quad g=\frac { 13 }{ 2 } ,f=\frac { -1 }{ 2 } ,c=3\)
The condition to represent pair of lines is abc + 2fgh - af² - bg² - ch² = 0
\((12)(k)(3)+2\left( \frac { -1 }{ 2 } \right) \left( \frac { 13 }{ 2 } \right) \left( \frac { 7 }{ 2 } \right) -12\left( \frac { 1 }{ 4 } \right) -k\left( \frac { 169 }{ 4 } \right) -3\left( \frac { 49 }{ 4 } \right) =0\)
\(\Rightarrow 36k-\frac { 91 }{ 4 } -3-\frac { 169k }{ 4 } -\frac { 147 }{ 4 } =0\)
\(\Rightarrow \frac { 144k-91-12-169k-147 }{ 4 } =0\)
\(\Rightarrow -25k-250=0\times 4=0\)
\(\Rightarrow -25k=250\)
\(\Rightarrow k=-10\)

The two lines are x (2 + 6k) + y(3 - k) + 4 + 12k = 0
7x + 5y - 4 = 0
Let m₁ and m₂ be the slopes of the given lines.
Then m₁=\(-\frac { { \text {Co-efficient of y} } }{ \text {Co-efficient of y }} =\frac { -(2+6k) }{ 3-k } \)
\(\\ and\ { m }_{ 2 }=\frac { -7 }{ 5 } \)
Since the given lines are perpendicular,
\(-\left( \frac { 2+6k }{ 3-k } \right) \left( \frac { -7 }{ 5 } \right) =-1\)
\(\Rightarrow 7(2+6k)=-5(3-k)\)
\(\Rightarrow 14+42k=-15+5k\)
\(\Rightarrow 37k=-29\)
\(\Rightarrow k=\frac { -29 }{ 37 } \)

The given lines are
2x - 5y + 3 = 0
5x - 9y + \(\lambda \) = 0
x - 2y + 1 = 0
The condition for the lines to be concurrent is
\(\left| \begin{matrix} 2 & -5 & 3 \\ 6 & -9 & \lambda \\ 1 & -2 & 1 \end{matrix} \right| =0\quad \)
Expanding along R₁ , we get
\(2\left| \begin{matrix} -9 & \lambda \\ -2 & 1 \end{matrix} \right| +5\left| \begin{matrix} 5 & \lambda \\ 1 & 1 \end{matrix} \right| +3\left| \begin{matrix} 5 & -9 \\ 1 & -2 \end{matrix} \right| =0\)
\(\Rightarrow 2(-9+2\lambda )+5(5-\lambda )+3(-10+9)=0\)
\(\Rightarrow -18+4\lambda +25-5\lambda -30+27=0\)
\(\Rightarrow -\lambda +4=0\Rightarrow \lambda =+4\)

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