Let P (x₁,y₁) be any point on the locus and  A (-1,1) and B (2,3) are the two given points
Area Of triangle  \(\triangle\)APB  = 8sq. units
= \(\frac { 1 }{ 2 } \)[x₁ (y₂ -y₃) + x₂ (y₃ - y₁)+ x₃ + (y₁ - y₂)]
\(\Rightarrow\)  \(\frac { 1 }{ 2 } \) [x₁ ( 1 - 3) + (-1) (3 - y₁) + 2 (y₁ - 1)] = 8
x₁ ( 1 - 3) + (-1) (3 - y₁) + 2 (y₁ - 1) = 16
\(\Rightarrow\)  -2x₁ -3 +y₁ +2y₁ - 2 - 16 = 0
\(\Rightarrow\)  -2x₁ + 3y₁ + 21 = 0
Locus of P(x₁, y₁) is 2x - 3y + 21 = 0

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

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

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.

let p(x₁,y₁) be any point on the locus ​\(\therefore\)​  \(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ \left( { y }_{ 1 }-0 \right) }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
\(\Rightarrow\)\(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ y }_{ 1 }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
Squaring  both sides, (x₁ - 4)² + y₁² = \(\frac{1}{4}\)(x₁ - 16)²
⇒ 4[x₁² + 16 - 8 x₁ + y₁²] = x₁² - 32x₁ + 256
⇒ 3x₁² + 4y₁² - 192 = 0
Locus of (x₁, y₁) is 3x² + 4y² = 192

Given lines are (b - c) x + (c - a) y + (a - b) = 0
(c - a) x + (a - b) y + (b - c) = 0
(a - b) x + (b - c) y + (c - a) = 0.
The condition for the given lines to be concurrent is
\(\left| \begin{matrix} b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a \end{matrix} \right| =0\)
Applying the elementary gains formation C₁⟶C₁+C₂+C₃
We get \(\left| \begin{matrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{matrix} \right| =0\)
Expanding along C₁ we get
\(\left| \begin{matrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{matrix} \right| =0\)
Hence the given lines are Concurrent

Slope of the line 3x-y+5=0 is
\(\Rightarrow \quad { m }_{ 2 }=-\frac { co-efficient\quad of\quad x }{ co-efficient\quad of\quad y } =\frac { -3 }{ -1 } =3\)
Let m₁ = m and \(\theta =45\)
\(\therefore \quad tan\theta =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\Rightarrow \quad tan\quad 45=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1|1+3m|=|m-3|\)
\(\Rightarrow \quad 1+3m=\pm (m-3)\)
\(\Rightarrow \quad 1+3m=m-3\quad or\quad 1+3m=-m+3\)
\(\Rightarrow\)   2m = - 4 or  4m = 2
\(\Rightarrow \)  m = - 2  or  \(m=\frac { 2 }{ 4 } =\frac { 1 }{ 2 }​​​​​​​\)
\(\therefore \)   m =  - 2 or \(\frac { 1 }{ 2 } \)

The given line is 2x + y = 1
Any line parallel to this is 2x + y + k = 0
Since this passes through (0,0),2(0) + 0+ k = 0 \(\Rightarrow\) k = 0
\(\therefore\)  Equation of parallel line is 2x + y = 0 .....(1)
Any line perpendicular to 2x + y = 1 is x - 2y + k₁ = 0
Since this passes through (0, 0), k₁ = 0
\(\therefore\)  Perpendicular line is x - 2y = 0 ......(2)
\(\therefore\) Required combined equation is
(2x + y)(x - 2y) = 0
\(\Rightarrow\) 2x² - 4xy + xy - 2y² = 0
\(\Rightarrow\) 2x² - 3xy - 2y² = 0.

Let (3, -4) and radius = Let C(3, -4) be the centre of the circle. If the line 5x + 12y - 12 = 0 touches the circle at P, then radius = perpendicular distance from p to the line 5x + 12y - 12 = 0.
\(\therefore r=\left| \frac { 5(3)+12(-4)-12 }{ \sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } } \right| =\frac { 45 }{ 13 } \)
\(\therefore\) Equation of the circle with centre 
(3,-4) and radius= \(\frac { 45 }{ 13 } \quad is\)
(x - 3)² + (y + 4)² = \({ \left( \frac { 45 }{ 13 } \right) }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+6x+9+{ y }^{ 2 }+8y+16=\frac { 2025 }{ 169 } \)
\(\Rightarrow \) 169 x² + 1014 x + 1521 + 169y² + 1352 y + 2704 = 2025
\(\Rightarrow \) 169 x² + 169 y² + 1014x + 1352y + 2200 = 0

4x² + 12y +9y² - 6x - 9y + 2 = 0
a = 4, 2h = 12
b = 9, h = 6
h² - ab = 36 - 36 = 0
Hence the given equation represents a pair of parallel straight lines.
To find separate equations
\(4 x^2+12 x y+9 y^2=(2 x+3 y)^2\)
\(4 x^2+12 x y+9 y^2-6 x-9 y+2=0\)
\((2 x+3 y)^2-3(2 x+3 y)+2=0\)
Let 2x + 3y = z
\(z^2-3 z+2=0\)
(z - 1) (z - 2) = 0,
Z - 1 = 0,   z - 2 = 0
2x + 3y -1 = 0, 2x + 3y - 2 = 0
Separate equations are 2x + 3y 1 = 0
2x + 3y - 2 = 0

3x² - 5xy - 2y² + 17x + y + 10 = 0
Comparing this with ax²+ 2hxy + by²+ 2gx + 2fy + c = 0
We get a=3, 2h = -5, b = -2, 
\(h=\frac { -5 }{ 2 }\)
Let \(\theta\) be the angle between the pair of straight lines then
tan \(\theta\) \(=\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \)
\(\Rightarrow tan\quad \theta =\frac { \pm 2\sqrt { \frac { 25 }{ 4 } +6 } }{ 1 } \)
\(\theta=\tan ^{-1}\left[\not 2\left(\frac{7}{\not 2}\right)\right]=\tan ^{-1}\)

Equation of circle is (x-h)² + (y - k)² = r²
(h, k) = (3, 5), r = 5
(x - 3)² + (y - 5)² = 5²
\(\Rightarrow\)  x² - 6x + 9 + y² - 10y + 25  = 25 
\(\Rightarrow\)  x² + y² - 6x - 10y + 9 = 0

Equation of circle is x² + y² = r²
r = 2
\(\Rightarrow\)  x² + y² = 4
\(\Rightarrow\)  x² + y² - 4 = 0

The given parabola x²=6y is of the form x²=4 ay where 4a=6 \(\Rightarrow a=\frac { 6 }{ 4 } \Rightarrow a=\frac { 3 }{ 2 } \)
Focus is (0, a)=\(\left( 0,\frac { 3 }{ 2 } \right) \)
Vertex is (0, 0)
Equation of the directrix is y=-a \(\Rightarrow\) \(y=\frac { -3 }{ 2 } \Rightarrow 2y+3=0\)
Length of the latus section = 4a=6 units.

On x axis y = 0
\(3 x=1 \Rightarrow x=\frac{1}{3}\)

\((x+a)^2+(y+b)^2=a^2-b^2\)
\(x^2+a^2+2 a x+y^2+b^2+2 b y=a^2-b^2\)

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

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