The supply price relation is given by
\({ x }^{ 2 }=5p-15\) 
= \(5(p-3)\) 
\(\Rightarrow { X }^{ 2 }aP\)  where X = x and \(P=p-3\) 
\(\\ \therefore \) the supply curve is a parabola whose vertex is \(\left( X=0,\ P=0 \right) \) 
i.e., The supply curve is a parabola whose vertex is (0, 3)

Here \({ r }^{ 2 }=25\Rightarrow r=5\) 
Parametric equations are \(x=rcos\theta ,y=rsin\theta \) 
\(\Rightarrow x=5cos\theta ,y=5sin\theta ,0\le \theta \le 2\pi \)

Equation of circle is \({ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }\) 
Here r = 3
i.e equation of circle is \({ x }^{ 2 }+{ y }^{ 2 }=9\)

Equation of circle is
\(\left( x-h \right) ^{ 2 }+\left( y-k \right) ^{ 2 }={ r }^{ 2 }\) 
Here \(\left( h,k \right) =(3,-1)\) and r = 4
Equation of circle is
\(\left( x-3 \right) ^{ 2 }+\left( y+1 \right) ^{ 2 }=16\)
\({ x }^{ 2 }-6x+9+{ y }^{ 2 }+2y+1=16\)
\({ x }^{ 2 }+{ y }^{ 2 }-6x+2y-6=0\)

We know that the acute angle \(\theta \) between two lines with slopes m₁ and m₂ is given by
\(tan\theta =\left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \) 
Given \({ m }_{ 1 }=3\ and\ \theta =\cfrac { \pi }{ 4 } \) 
\(\therefore tan\cfrac { \pi }{ 4 } =\left| \cfrac { 3-{ m }_{ 2 } }{ 1+3{ m }_{ 2 } } \right| \)
\(1=\left( \cfrac { 3-{ m }_{ 2 } }{ 1+3{ m }_{ 2 } } \right)\)
\(1+{ 3m }_{ 2 }=3-{ m }_{ 2 }\)
\( \Rightarrow { m }_{ 2 }=\cfrac { 1 }{ 2 } \) 
Hence the slope of the other line is \(\frac { 1 }{ 2 } \).

Given line is x - y + 5 = 0
Perpendicular distance of the given line from P(2, 2) is
= \(\left| \cfrac { 2-2+5 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \right| \) 
= \(\left| \cfrac { 5 }{ \sqrt { 2 } } \right| =\cfrac { 5 }{ \sqrt { 2 } } \) 
Distance of (0,0) from the given line
= \(\left| \cfrac { 5 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \right| =\left| \cfrac { 5 }{ \sqrt { 2 } } \right| =\cfrac { 5 }{ \sqrt { 2 } } \) 
The given line is equidistance from origin and (2, 2).

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.

Equation of the parabola is x²+y=6x-14
⇒ x²-6x=-y-14
⇒ x²-6x+9=-y-14+9 (Adding 9 both sides)
⇒ (x-3)²=-y-5
⇒ (x-3)²=-1(y+5)
⇒ x²=-y where X=x-3, Y=y+5

The given equation is
y²=4x+4y
⇒ y²-4y=4x
⇒ y²-4y=4x
⇒ y²-4y+4=4x+4 (Adding 4 on both sides)
⇒ (y-2)²=4(x+1)
⇒ y²=4 where X=x+1 ⇒ Y=y-2

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

x + y = 6 .....(1)  
x + 2y = 4  .....(2)
On solving (1) and (2) we get the centre of the circle
C(-h, - k) = (8, - 2)
Equation of circle is \((x-h)^2+(y-k)^2=r^2\)
\((x-8)^2+(y+2)^2=r^2\)
It passes through (2, 6)
\((2-8)^2+(6+2)^2=r^2\)
\(36+64=r^2 \Rightarrow r^2=100\)
Required equation is \((x-8)^2+(y+2)^2=100\)
\(x^2-16 x+64+y^2+4 y+4=100\)
\(x^2+y^2-16 x+4 y-32=0\)

c = 16\(\pi\)
\(\Rightarrow\)  2\(\pi\)r  = 16 \(\Rightarrow\) r = 8
∴ Equation of the circle is \((x-h)^2+(y-k)^2=r^2\)
( x + 3)² + (y + 2)² = 8²
\(\Rightarrow\) x² + 6x + 9 +y² + 4y + 4 = 64
\(\Rightarrow\) x² + y² + 6x + 4y + 13 - 64 = 0
\(\Rightarrow\) x² + y² + 6x + 4y - 51 = 0

(x + 2) ( x - 5) + (y -2 ) ( y -1) = 0
\(\Rightarrow\) x² -5x + 2x - 10 + y² - y - 2y + 2 = 0
\(\Rightarrow\)  x² + y² - 3x - 3y - 8 = 0
here 2g  = -3    \(\Rightarrow\)  \(g=-\frac { 3 }{ 2 }\)
2f  = -3   \(\Rightarrow\)   \(f=-\frac { 3 }{ 2 } \)
and C = -8
Center of the circle (-g, -f) = \(\left( \frac { 3 }{ 2 } ,\frac { 3 }{ 2 } \right) \)
A radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
Radius of the circle is  \(= \sqrt { \frac { 9 }{ 4 } +\frac { 9 }{ 4 } +8 }\)
\(=\sqrt{\frac{18}{4}+8}=\sqrt{\frac{9}{2}+8}\)
\(r=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}} \text { units }\)

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

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