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 
\({ 2x }^{ 2 }+{ 2y }^{ 2 }+4x+4y-1=0\)

Let m₁ and m₂ be the slopes of the pair of straight lines
\({ ax }^{ 2 }+2hxy+{ by }^{ 2 }=0\)
\(\therefore { m }_{ 1 }+{ m }_{ 2 }=-\cfrac { 2h }{ b } \quad { m }_{ 1 }{ m }_{ 2 }=\cfrac { a }{ b } \) 
It is given that one slope is twice the other, so let \({ m }_{ 2 }=2{ m }_{ 1 }\) 
\(\therefore { m }_{ 1 }+2{ m }_{ 1 }=-\cfrac { 2h }{ b } \quad { m }_{ 1 }.{ 2m }_{ 1 }=\cfrac { a }{ b } \)
\(\therefore { m }_{ 1 }=-\cfrac { 2h }{ 3b } \quad and\quad 2{ m }_{ 1 }^{ 2 }=\cfrac { a }{ b } \)
\(\Rightarrow 2\left( -\cfrac { 2h }{ 3b } \right) ^{ 2 }=\cfrac { a }{ b } \)
\( \Rightarrow \cfrac { { 8h }^{ 2 } }{ { 9b }^{ 2 } } =\cfrac { a }{ b } \)
\( \Rightarrow { 8h }^{ 2 }=9ab\)

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,
\(abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0\)
\(i.e., \ 4k+\cfrac { 675 }{ 2 } -162\cfrac { 225 }{ 2 } -\cfrac { 25 }{ 4 } k=0\)
\( \Rightarrow 16k+1350-648-450-25k=0\)
\(\Rightarrow 9k=252\therefore k=28\)

Given lines 3x - 4y - 13 = 0...(1)
8x - 11y = 33  ...(2)
2x - 3y - 7 = 0 ...(3)
Conditon for concurrent lines is
\(\left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{matrix} \right| =0\) 
i.e.,\(\left| \begin{matrix} 3 & -4 & -13 \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{matrix} \right| =3\left( 77-99 \right) +4\left( 56+44 \right) -13\left( -24+22 \right) \) 
= -66 + 40 + 26 = 0
\(\Rightarrow \) Given lines are concurrent. To get the point of concurrency solve the equations (1) and (3)

When y = 5 from (2) 8x = 88
x = 11
Point of concurrency is (11, 5)

Y = -x² + 10 x - 15
\(\Rightarrow\) x² - 10x = - y - 15
\(\Rightarrow\) (x- 5)² = -y - 15 + 25
\(\Rightarrow\) (x- 5)² = -y + 10 
\(\Rightarrow\) (x - 5)² = (-y - 10)
The best time to end the project is when x = 5 months.