(y - 2)² = 4(x - 1).
\(\Rightarrow\) y² = 4x where X = x - 1 and Y = y - 2. \(\Rightarrow\) 4a = 4 \(\Rightarrow\) a = 1

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

Let x and Y represent the number of units produced and the cost of production respectively. By the given data.
x₁ (500)  y₁ (6000)
x₂ (1000)  y₂ (9000)
Equation of straight line is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \frac { y-6000 }{ 9000-6000 } =\frac { x-500 }{ 1000-500 } \)
\( { y-6000 }{ } =\frac { 3000 }{ 500 } (x - 500)\)
\(\Rightarrow\) y - 6000 = 6(x - 500)
\(\Rightarrow\) y - 6000 = 6x - 3000
\(\Rightarrow\) y = 6x + 3000

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

Compare the equation
12x² - 10xy + 2y² + 14x - 5y + 2 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 12, 2h = -10, b = 2, 2g = 14, 2f = -5
\(h=-5\quad g=7\quad f=-\frac { 5 }{ 2 } ,c=2\)
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 12 & -5 & 7 \\ -5 & 2 & \frac{-5}{2} \\ 7 & \frac{-5}{2} & 2 \end{array}\right|\)
\(=12\left(4-\frac{25}{4}\right)+5\left(-10+\frac{35}{2}\right)+7\left(\frac{25}{2}-14\right)\)
\(=48-75-50+\frac{175}{2}+\frac{175}{2}-98\)
= -175 + 175 = 0
Hence the given equations represent a pair of straight lines.
To find separate equation
\(12 x^2-10 x y+2 y^2 =12 x^2-6 x y-4 x y+2 y^2 \)
\(=6 x(2 x-y)-2 y(2 x-y) \)
\(=(6 x-2 y)(2 x-y)\)
\(12 x^2-10 x y+2 y^2+ 14 x-5 y+2 =(6 x-2 y+l)(2 x-y+\mathrm{m})\)
Comparing the coefficient of x and y
14 = 6m + 2l
divided by 2
7 = 3m + l .........(1)
-5 = -2m - l .......(2)
Solving (1) and (2) we get m = 2, 1 = 1
The separate equations are
6x - 2y + 1 = 0
2x - y + 2 = 0