\(4 a=1 \Rightarrow a=\frac{1}{4}\)
Equation x = a
\(x=\frac{1}{4}\)

\(2 n r=8 \pi\)
r = 4

Equation of x axis y = 0, equation of y axis x = 0
\(\therefore\) combined equation xy = 0

C(-g, -f)

4a = 5

\(r=\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3\)

\(a+b=0 \Rightarrow k-2=0\)

(parallel line)

Let p(x,y) be any point on the parabola whose focus is F(-3, 2) and the directrix is x + y - 4 = 0.
Draw PM perpendicular to x + y - 4 = 0
Then FP = PM  \(\Rightarrow\) FP² = pm²
\(\Rightarrow { (x+3) }^{ 2 }+{( y-2) }^{ 2 }={ \left[ \frac { x+y-4 }{ \sqrt { 1+1 } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+6x+9+{ y }^{ 2 }-4y+4=\frac { { x }^{ 2 }+{ y }^{ 2 }+16+2xy-8x-8y }{ 2 } \)
\(\Rightarrow\) 2(x² + y² + 6x - 4y + 13) = x² + y² + 2y - 8x - 8y + 16
\(\Rightarrow\) x² + y² - 2xy + 20x + 10 = 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 coordinates 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 coordinates 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 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

(h, k) = (2 , 3)
Equation of circle is \((x-h)^2+(y-k)^2=r^2\)
(x - 2)² + ( y - 3)² = r²
It passes through (1 , 4)
(1 - 2)² + (4 -3)² = r²
1 + 1 = r² \(\Rightarrow\) r² = 2
Required equation
(x - 2)² + ( y - 3)² = 2
\(\Rightarrow\) x² - 4x + 4+ y² - 6y + 9 = 2
\(\Rightarrow\) x² + y² - 4x - 6y + 11 = 0

5x² + 5y² +4x - 8y - 16 = 0
[Divide by 5]
x² + y² +  \(\frac { 4 }{ 5 } x-\frac { 8 }{ 5 } y-\frac { 16 }{ 5 } =0\)
Here 2g  = \(\frac { 4 }{ 5 } \) \(\Rightarrow\)  \(g=+\frac { 2 }{ 5 } \)
2f = \(-\frac { 8 }{ 5 } \)     \(\Rightarrow\) \(f=-\frac { 4 }{ 5 } \)
and c = \(-\frac { 16 }{ 5 } \)
Center of the circle is (-g, -f) \(\Rightarrow \) \(\left( -\frac { 2 }{ 5 } ,\frac { 4 }{ 5 } \right) \)
Radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\(\Rightarrow\) r = \(\sqrt { \frac { 4 }{ 25 } +\frac { 16 }{ 25 } +\frac { 16 }{ 5 } } =\sqrt { \frac { 20 }{ 25 }+ { \frac { 16 }{ 5 }} } \)
\(\Rightarrow\) \(r=\sqrt { \frac { 20 }{ 5 } } \) = \(\sqrt { 4 } \) = 2 units

x² + y² - 22x - 4y + 25 = 0
2g  = - 22   \(\Rightarrow \)  g = -11
2f  = -4   \(\Rightarrow\) f = -2
c = 25 
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0\)    
Center of the circle is (-g, -f) = ( 11, 2)
Radius of the circle is  \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { \left( -11 \right) ^{ 2 }+({ -2 })^{ 2 }-25 } \)
 \(r= \sqrt { 121+4-25 } =\sqrt { 100 } \) = 10 units

x² + y² = 16
\(\therefore\)  Centre is (0,0), r² = 16
r = 4 units

Since the parabola is symmetric about X-axis and has its vertex at (0,0), its equation will be of the form y² = 4ax or y² = -4ax.
But the parabola passes through (2, 3) which is in the I quadrant, its equation will be of the form y² = 4ax, which is open rightward.
Substituting (2,3) in y² = 4ax, we get
9 = 4a(2) ⇒ 8a = 9
⇒ a = \(\frac{9}{8}\)
\(\therefore\) Equation of the parabola is y² = 4(\(\frac{9}{8}\))x
⇒ y² = \(\frac{9}{2}\)x
⇒ 2y² = 9x ⇒ 2y² - 9x = 0

Equation of circle be x² + y² + 2gx + 2fy + c = 0 ...(1)
It passes through (1, 0)
1 + 0 + 2g + 0 + c = 0   \(\Rightarrow\)  2g + c = -1..(2)
The circle passes through (0, 1)
0 + 1 + 0 + 2f + c = 0 \(\Rightarrow\)  2f + c = -1 ...(3)
centre (-g, -f) lies on x + y = 1
-g - f = 1  ...(4)
Solving (1), (2) and (3) we get
\(g=-\frac { 1 }{ 2 } , f=-\frac { 1 }{ 2 } \) c = 0
Equation of circle is
\(\therefore \)x² + y² + 2 \(\left( -\frac { 1 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) y+0=0\)
\(\Rightarrow\) x² + y² - x - y = 0 

Equation of the circle be x² + y² + 2gx + 2fy + c = 0  ..(1)
It passes through (0, 1)
0 + 1 + 0 + 2f + c = 0   \(\Rightarrow \) 2f + c = -1 ..(2)
The circle passes through (4, 3)
16 + 9 + 8g + 6f + c = 0  \(\Rightarrow \) 8g + 6f + c = -25 ...(3)
The circle passes through (1,-1)
1 + 1 + 2g - 2f + c = 0  \(\Rightarrow \) 2g - 2f + c = -2 ....(4)
Solving (1), (2) and (3) we get
c = 1   \(\Rightarrow\)  g = -5/2    \(\Rightarrow\)  f = -1
Equation of circle is
x² + y² + 2 \(\left( -\frac { 5 }{ 2 } \right) \)x + 2(-1)y + 1 = 0
x² + y² - 5x - 2y + 1 = 0