Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

Here A = 0, B = 0, C = 1, D = 4, E = 3, F = 4
B = 0, A = 0 either A or C is 0.
Hence, the given equation represents a parabola.

Here A = 1, B = 0, C = 1, D = 1, E = -1
Here A = C and B = 0 there i no xy term.
Hence, the given equation represent a circle.

Here A = 3, B = 0, C = 3, D = -4, E = 3 and F = 10
A = C and B = 0 (No xy term)
Hence, the given equations represents a circle.

Taking (x₁, y₁) as (2, 3), we get
x₁² + y₁² + 2gx₁+ 2fy₁+ c = 2² + 3² − 6 × 2 − 8 × 3 + 12 
= 4 + 9 - 12 - 24 + 12
= -11\(<\)0.
Therefore the point (2, 3) lies inside the circle, by theorem.

Equation of the circle in standard form is (xr − h)² + (y − k)² = r²
\( \Rightarrow (x-(-3))^{2}+(y-(-4))^{2} =3^{2} \)
\( \Rightarrow (x+3)^{2}+(y+4)^{2} =3^{2} \)
\( \Rightarrow x^{2}+y^{2}+6 x+8 y+16 =0 .\)

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Parabola is open left and axis of symmetry as x-axis and vertex (0, 0)  
Then the equation of the required parabola is
(y - 0)² = -4\(\sqrt { 2 } \) (x - 0)
y² = -4\(\sqrt { 2 } \) x

Given that equation of the circle is
x² + y² = 16
⇒ a² = 16
and equation of the tangent is
y = \( 2\sqrt { 2 } \)x + c
⇒ m = \( 2\sqrt { 2 } \)  and c = c
[∵ y = mx + c is the tangent]
The condition for the line y = mx + c is a tangent to the circle x² +y² = a² is
c² = a² (1 + m²)
⇒ c² = 16(1 + \(\left( { \left( 2\sqrt { 2 } \right) }^{ 2 } \right) \)
⇒ c² = 16(1 + 8)
⇒  c² = 16(9)
⇒ c = ± 4(3)
⇒ ±12

Since the circles touch both the axes. Its equation will be
(x + a)² + (y + a)² = a² ...............(1)
It passes through (-4, -2)
∴ (-4 + a)² + (-2 + a)² = a²
\(16+\not a^{2}+8 a+4+a^{2}+4 a=\not a^{2}\)
⇒ a² + 12a + 20 = 0
⇒ (a + 10)(a + 2) = 0
a = -10 or -2

Case (i):
When a = -10, (1) becomes
(x + 10)² + (y + 10)2 = 10²
\(\Rightarrow x^{2}+\not 100+20 x+y^{2}+\not 100+20 y=160\)
⇒  x² + y²+ 20x + 20y + 100 = 0
Case (ii):
When a = -2, (1) becomes
⇒ (x + 2)² + (y + 2)² = 2²
\(x^{2}+4 x+4+y^{2}+4 y+\not 4 = \not 4\)
x² + y²+ 4x + 4y + 4 = 0
Hence, equation of the circles are
x² + y²+ 4x + 4y + 4 = 0
or  x² + y²+ 20x + 20y + 100 = 0

Given centre is (2, -1) and passing through the point (3, 6)
∴ r = distance between (2, -1) and (3, 6)
= \(\sqrt { { (2-3) }^{ 2 }+(-1{ -6) }^{ 2 } } \)
= \(\sqrt { { (-1) }^{ 2 }+({ -7) }^{ 2 } } \)
= \(\sqrt { 1+49 } =\sqrt { 50 } \)
∴ Equation of the circle is
(x - h)² + (y - k)² = r²
(x−2)²+(y+1)² = \({ (\sqrt { 50 } })^{ 2 }\)
(x−2)²+(y+1)² = 50

Let O₁ O₂ be the centres of the two semi circular vents.
First vent with centre O₁ (12, 0) and radius r = 10 yields equation to first semicircle as
(x−12)²+(y− 0)² = 10²
\(\Rightarrow\) x²+y²−24x + 44 = 0, y > 0
Second vent with centre O₂ (34, 0) and radius r = 10 yields equation to second vent as
(x−34)²+ y² = 10² 
x²+y²− 68x  + 1056 = 0, y > 0 

C(2, 1) is the centre and 3x + 4y + 10 = 0 cuts a chord AB on the circle. Let M be the midpoint of AB,
then AM = BM = 3. Now BMC is a right triangle.
So, we have CM = \(\frac { \left| 3\left( 2 \right) +4\left( 1 \right) +10 \right| }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =4\)
By Pythagoras theorem BC² = BM² + MC² = 3² + 4² = 25
BC = 5 = radius
Equation of the required circle is
(x−2)²+(y−1) = 5²
x²+y²−4x−2y−20 = 0 .

Equation of ellipse is
x²+ 4y² = 32
\(\frac { { x }^{ 2 } }{ 32 } + \frac { { y }^{ 2 } }{ 8 } =1\)
a² = 32, b² = 8
\(a=4\sqrt { 2 } ,b=2\sqrt { 2 } \)
Equation of tangent at \(\theta =\frac { \pi }{ 4 } \) is \(\frac { xcos\frac { \pi }{ 4 } }{ 4\sqrt { 2 } } \frac { ysin\frac { \pi }{ 4 } }{ 2\sqrt { 2 } } =1\) 
\(\frac { x }{ 8 } +\frac { y }{ 4 } =1\)
x+2y−8 = 0.
Equation of normal is \(\frac { 4\sqrt { 2X } }{ cos\frac { \pi }{ 4 } } -\frac { 2\sqrt { 2Y } }{ sin\frac { \pi }{ 4 } } =32-8\)
That is 8x-4y = 24 
2x-y-6 = 0
Aliter: 
At, \(\theta =\frac { \pi }{ 4 } \)
\((a\ cos \theta ,b\ sin \theta )=\left( 4\sqrt { 2 }\ cos\frac { \pi }{ 4 } ,2\sqrt { 2 }\ sin\frac { \pi }{ 4 } \right) \)
= (4, 2)
∴ Equation of tangent at \(\theta =\frac { \pi }{ 4 } \) is same at (4, 2) 
Equation of tangent in cartesian form is \(\frac { { xx }_{ 1 } }{ { a }^{ 2 } } +\frac { { yy }_{ 1 } }{ { b }^{ 2 } } =1\)
x+2y−8 = 0
Slope of tangent is -\(\frac { 1 }{ 2 } \)
Slope of normal is 2 Equation of normal is  y - 2 = 2(x -4) 
y−2x+6 = 0 

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)
This is an equation of the ellipse.
a² = 2⁵ and b² = 9 and c² = a² - b²
⇒ c² = 25 - 9 = 16 ⇒ c = 4
(a) Center is (0, 0) ⇒ h = 0, k = 0
(b) foci are (h - c, k), (h + c, k)
⇒ (0 - 4, 0), (0 + 4, 0)
⇒ (-4, 0) and (4, 0)
(c) Vertices are (h - a, k) and (h + a, k)
⇒ (0 - 5, 0) and (0 + 5, 0)
⇒ (-5, 0) and (5, 0)
(d) Directrices are x = \(\pm \frac { a }{ e } \)
⇒ x = \(\pm \frac { 5 }{ e } \)
\(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
∴ Directrice are x = \(\pm \frac { 5 }{ \frac { 4 }{ 5 } } \Rightarrow x=\pm \frac { 25 }{ 4 } \)

Equation of the circle is x² + y² − 6x + 6y − 8 = 0
∴ Equation of the tangent at (x₁, y₁) is
xx₁ + yy₁ - \(\frac 62\) (x + x₁) + \(\frac 62\) (y + y₁) - 8 = 0
Given (x₁, y₁) is (2, 2)
Equation of the tangent at (2, 2) is
x(2) + y(2) - 3(x + 2) + 3(y + 2) - 8 = 0
\(\Rightarrow 2 x-2 y-3 x+\not 6+3 y+\not 6-8=0\)
⇒ -x + 5y - 8 = 0
⇒ x − 5y + 8 = 0
Equation of the normal is
yx₁ - xy₁ + g(y - y₁) - f(x - x₁) = 0
⇒ y(2) -x(2) - 3(y - 2) -3(x - 2) = 0
∵ 2g = -6
⇒ g = -3
2f = 6
⇒ f = 3
⇒ 2y - 2x - 3y + 6 - 3x + 6 = 0
⇒ -5x - y + 12 = 0
Another method for Normal:
Equation of tangent is perpendicular to normal
x - 5y + 8 = 0
Perpendicular equation be 5x + y + k = 0
At (2, 2)
10 + 2 + k = 0
k = -12
Therefore 5x + y - 12 = 0 be equation of normal.

(2) Equation of directrix is x = h + a

(4) Transverse axis parallel to y-axis

(4) (a cos θ, b sin θ) ; 1, 2, 3 are parametric form of a circle