The given equation is
\(\left( a-2 \right) { x }^{ 2 }+{ by }^{ 2 }+\left( b-2 \right) xy+4x+4y-1=0\) 
As per conditions noted above,
(i) coefficient of \(xy=0\Rightarrow b-2=0\ \therefore b=2\) 
(ii) coefficient of x² = coefficient of y²
\(\Rightarrow a-2=b\)
\(a-2=2\Rightarrow a=4\) 
\(\therefore \) Resulting equation of circle is 
\({ 2x }^{ 2 }+{ 2y }^{ 2 }+4x+4y-1=0\)

The given equation is \({ 4x }^{ 2 }-12xy+{ 9y }^{ 2 }+18x-27y+8=0\) 
Here a = 4, b = 9 and h = –6
\({ h }^{ 2 }-ab=36-36=0\) 
Hence the given equation represents a pair of parallel straight lines
Now \({ 4x }^{ 2 }-12xy+{ 9y }^{ 2 }=\left( 2x-3y \right) ^{ 2 }\) 
Consider, \({ 4x }^{ 2 }-12xy+{ 9y }^{ 2 }+18x-27y+8=0\)
\( \Rightarrow \left( 2x-3y \right) ^{ 2 }+9\left( 2x-3y \right) +8=0\) 
Put 2x – 3y = z
\({ z }^{ 2 }+9z+8=0\)
\(\left( z+1 \right) \left( z+8 \right) =0\)
\(z+1=0\quad z+8=0\) 
2x–3y+1 = 0  2x–3y+8 = 0
Hence the separate equations are
\(\\ 2x-3y+1=0\) and \(2x-3y+8=0\)

Here \(a=2,b=2,h=\cfrac { 5 }{ 2 } ,g=\cfrac { 15 }{ 2 } ,f=9,c=k\) 
The given line represents a pair of straight lines if,
\(abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0\)
\(i.e., \ 4k+\cfrac { 675 }{ 2 } -162\cfrac { 225 }{ 2 } -\cfrac { 25 }{ 4 } k=0\)
\( \Rightarrow 16k+1350-648-450-25k=0\)
\(\Rightarrow 9k=252\therefore k=28\)

The given equation is \({ x }^{ 2 }+4xy+{ y }^{ 2 }=0\)
Here a = 1, b = 1 and h = 2
If \(\theta \) is the angle between the given straight lines, the
\(\theta ={ tan }^{ -1 }\left[ \left| \cfrac { 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \right| \right] \)
\(={ \tan }^{ -1 }\left[ \left| \cfrac { 2+\sqrt { 4-1 } }{ 2 } \right| \right]\)
\(= { \tan }^{ -1 }\left( \sqrt { 3 } \right) \)
\(\theta =\cfrac { \pi }{ 3 } \)

Let the equation of the circle be
 x² + y² + 2gx + 2fy + c =0 ...(1)
Since (2, 3) lies on (1) we get
4 + 9 + 4g + 6f + c = 0
\(\Rightarrow \) 4g + 6f + c = -13..(2)
Since (-1, 4) lies on (1) we get
1 + 11 - 2g + 2f + c = 0
\(\Rightarrow \) -2g + 2f + c =-2....(3)
Also (-g₁-f) lies on x - 3y - 11 = 0
\(\Rightarrow \) -g + 3f = 11....(4)
(2) - (3) we get,
6g + 4f = -11
(4) \(\times\) 6 \(\rightarrow\) -6g + 18f = 66
22f = 55
\(f=\frac { 5 }{ 2 } \)
Substituting \(f=\frac { 5 }{ 2 } \) in (4) we get
\(=-g+\frac { 15 }{ 2 } =11\quad \Rightarrow -g=11-\frac { 15 }{ 2 } =\frac { 22-15 }{ 2 } =\frac { 7 }{ 2 } \)
\(\Rightarrow \quad g=\frac { 7 }{ 2 } \)
Substituting \(f=\frac { 5 }{ 2 } ,g=\frac { -7 }{ 2 } \) in (3) we get
\(-2\left( \frac { -7 }{ 2 } \right) +2\left( \frac { 5 }{ 2 } \right) +c=-2\)
7 + 5 + c = - 2   \(\Rightarrow\) c = -14
Equation of the required circle is
\({ x }^{ 2 }+{ y }^{ 2 }+2\left( \frac { -7 }{ 2 } \right) x+2\left( \frac { 5 }{ 2 } \right) y-14=0\)
\(\Rightarrow\)  x² + y² - 7x + 5y - 14 =0.

Equation of the circle is x² + y² - 2x + 6y - 15 = 0
2g = -2, 2f = 6
g = - 1, f = 3
Centre (-g, f) = (1, -3)
Let A(4, 1) and B(x, y) be the 2 ends of the diameter
Midpoint of AB \(=\left(\frac{4+x}{2}, \frac{1+y}{2}\right)=c(1,-3)\)
\(\frac{4+x}{2}=1, \ \frac{1+y}{2}=-3\)
\(4+x=2,1+y=-6\)
x = - 2, y = - 7
B(-2, -7) is the other end of the diameter

(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

Since the given line are concurrent
\(\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0\)
\(\left| \begin{matrix} 3 & 4 & -13 \\ 2 & -7 & 1 \\ a & -1 & -14 \end{matrix} \right| =0\)
\(\Rightarrow \) 3 (98 +1) - 4 (-28 -a ) -13 (-2 +7a) = 0
\(\Rightarrow \) - 297 +112 +4a +26 -91a = 0
\(\Rightarrow \) 435 =87a
\(\Rightarrow \) a =\(\frac { 435 }{ 87 } \)
\(\Rightarrow \) a = 5

The condition for concurrent line is
\(\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0\)
\(\Rightarrow \) \(\left| \begin{matrix} 1 & 1 & -4 \\ 3 & 0 & 2 \\ 3 & -3 & 16 \end{matrix} \right| =0\)
\(\Rightarrow \)  1 (0+6) -1 (48 -6) - 4 (-9 -0)
\(\Rightarrow \) 6 - 42 + 36 = 0
\(\Rightarrow \) 42 - 42 = 0
Hence,therefore the given lines are concurrent

\(h^2-a b=0\)
\(4-2 a=0 \Rightarrow a=2\)

4a = 5

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

Taking vertex of the parabola as reflector at origin, x-axis along the axis of the parabola, equation of the parabola is y² = 4ax
Given depth = 5 cm, diameter = 20 cm
\(\therefore\) (5, 10) lies on the parabola
\(\therefore\) 10² = 4a(5) ⇒ 100 = 20a ⇒ a = \(\frac{100}{20}\) = 5

\(\therefore\) Focus is (a, 0) = (5, 0) which is the mid-point of the given diameter

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 straight line is 3x + 2y + 7 = 0
At P(-1, -2) PT = -3 - 4 + 7 = 0
At Q(1, 0) QT = 3 + 0 + 7 = 10 > 0
At R (-3, -4) RT = -9 - 8 + 7 = - 10 < 0
P lies on the straight line, Q above the line and R below the line

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

The combined equation of the given straight lines is
\(\left( 2x+y-1 \right) \left( x+2y-5 \right) =0\)
\(i.e \ \ { 2x }^{ 2 }+xy-x+4xy+2{ y }^{ 2 }-2y-10x-5y+5=0\)
\( i.e \ \ { 2x }^{ 2 }+5xy+2{ y }^{ 2 }-11x-7y+5=0\)