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.

Equation of the circle is x² + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

Equation of the circle is
x² + y² + 6x - 4y + 4 = 0.
Here 2g = 6 ⇒ g = 3
2f  = -4 ⇒ f = -2 and c = 4
Centre is (-g, -f) ⇒ (-3, 2)
r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { 3 }^{ 2 }+(-2){ }^{ 2 }-{ 4 } } \)
= \(\sqrt { 9+4-4 } \)
= \(\sqrt { 9 } \)
= 3 unit

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.

A = 11, C = -25, D = -44, E = - 50, and F = -256
Here A ≠ C and A and C are of opposite signs. Hence, the given equation represents a hyperbola.

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, C = 2 and F = -14
A ≠ C and A and C are of the same sign.
Hence, the given equation represents an ellipse.

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.

Here A = 2, B = 0, C = -1, F = -7
Here A ≠ C and A and C are of opposite signs.
Hence the given equation represents a hyperbola.

Given that 2a = 36.18, 2b = 9.12 , we get
e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\frac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \) = \(\frac { \sqrt { { \left( \frac { 36.18 }{ 2 } \right) }^{ 2 }{ \left( \frac { 9.12 }{ 2 } \right) }^{ 2 } } }{ \frac { 36.18 }{ 2 } } \)
\(\frac { \sqrt { { \left( 18.09 \right) }^{ 2 }-{ \left( 4.56 \right) }^{ 2 } } }{ \left( 8.09 \right) } \approx0.97\)

Reducing 9x²-16y² = 144 to the standard form,
we have, \(\frac { { x }^{ 2 } }{ 16 }- \frac { { y }^{ 2 } }{ 9 } =1\)
With the transverse axis is along x-axis vertices are (−4, 0) and (4, 0); and c² = a²+b² = 16 + 9 = 25, c = 5  
Hence the foci are (−5, 0) and (5, 0) 

Equation of the circle is x² + (y + 2)² = 0
Compare with(x-h)²+(y-k)² = r²
h = 0, k = -2, r² = 0
Centre (h, k) = (0, -2)
radius is 0.

Given ends of diameter are (3, 4)(2, -7)
∴ Equation of the circle is
(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
⇒ (x - 3)(x - 2) + (y - 4)(y + 7) = 0
⇒ x² - 2x - 3x + 6 + y² + 7y - 4y - 28 = 0
⇒ x² + y² − 5x + 3y − 22 = 0

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 with end points of the diameter as (x₁, y₁₎ and (x₂, y₂) given in theorem is
(x−x₁)(x−x₂)+(y−y₁)(y−y₂) = 0
(x+4)(x−1)+(y+2)(y−1) = 0
x² + y² + 3x + y − 6 = 0 which is the required equation of the circle.

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

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

ae = distance between centre and focus
ae = \(\sqrt { { (8-2) }^{ 2 }-(1{ -1) }^{ 2 } } =\sqrt { { 6 }^{ 2 } } =6\) ...(1)
Also \(\frac { a }{ e } =\sqrt { (4{ -2) }^{ 2 }+(1{ -1) }^{ 2 } } =\sqrt { { 2 }^{ 2 } } =2\)
[∴ (4, 1) is a point on the directrix]
\((1) \times(2) \rightarrow a \not e \times \frac{a}{\not e}=6 \times 2\)
⇒ a² = 12
(1) ➝ a²e² = 36
12(e²) = 36
⇒ e² = 3
⇒ e = \(\sqrt { 3 } \)
Also, b² = a²(e² - 1) = 12(3- 1) = 12(2) = 24
∴ Equation of the hyperbola is
\(\frac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } -\frac { (y{ -k) }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { (x-2) }^{ 2 } }{ 12 } -\frac { (y{ -1) }^{ 2 } }{ 24 } =1\)

\(a\left( \frac { 3 }{ 2 } \right) =2\Rightarrow a\frac { 4 }{ 3 } \)
b² = a²(e² - 1)
\(\Rightarrow b^{2}=\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{16}{9}\left(\frac{9-4}{4}\right)=\frac{\not 16}{9} \times \frac{5}{4}\)
⇒ \({ b }^{ 2 }=\frac { 4\times 5 }{ 9 } =\frac { 20 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ \frac { 16 }{ 9 } } -\frac { { y }^{ 2 } }{ \frac { 20 }{ 9 } } =1\)
⇒ \(\frac { { 9x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 20 } =1\)

 Vertex is (1, -2) and focus is (4, -2). The parabola is right open.

In given data the parabola is open rightwards and symmetric about the line parallel to x-axis. Equation of parabola
a = distance between (1, -2) and (4, -2)
⇒ a = \(\sqrt { { (1-4) }^{ 2 }+(-2+{ 2) }^{ 2 } } \)
⇒ a = \(\sqrt { { (-3) }^{ 2 } } \) = 3
Equation of the parabola is
(y - k)² = 4a(x-h)
(y + 2)² = 4(3) (x - 1)
[∵ (h, k) is (1, -2)]
⇒ (y+2)² = 12 (x−1)

The parabola passes through (2, -3) and symmetric about y-axis.

Since the parabola passes through (2, -3) the parabola is open downward.
∴ Its equation is x² = -4ay ...(1)
Substitute the point (2, -3) we get,
22 = -49(-3)
⇒ 4 = 12a
\(\Rightarrow a=\frac{A}{\not \frac{12}{3}}=\frac{1}{3}\)
(1) becomes,
∴ x² = -4\(\left( \frac { 1 }{ 3 } \right) \)y
⇒ 3x² = −4y

Given Focus (4, 0) and directrix x = -4.
Since (4, 0) is the focus and the directrix is x = -4, the parabola is of the form y² = 4ax

Also a = 4
∴ Equation of the parabola is y² = 4(4)x
⇒ y² = 16x

y² = -8x
The given parabola is left open parabola
and 4a = 8 ⇒ a = 2
(a) Vertex is (0, 0)
⇒ h = 0, k = 0
(b) focus is (h - a, 0 + k)
⇒ (0-2, 0 + 0)
⇒ (-2, 0)
(c) Equation of directrix is x = h + a
⇒ x = 0 + 2 ⇒ x = 2
(d) Length of latus rectum is 4a = 8.

x² = 24y
The given parabola is open upward parabola and 4a = 24 ⇒ a = 6.
(a) Vertex is (0, 0)
⇒ h = 0, k = 0
(b) focus is (0 + h, a + k)
⇒ (0 + 0, 6 + 0) = (0, 6)
(c) Equation of directrix is y = k - a
⇒ y = 0 - 6 ⇒ y = -6
(d) Length of latus rectum is 4a = 24

y² = 16x
The given parabola is right open parabola and 4a = 16 ⇒ a = 4.
(a) Vertex is (0, 0)
⇒ h = 0, k = 0
(b) focus is (h + a, 0 + k)
⇒ (0 + 4, 0 + 0) = (4, 0)
(c) Equation of directrix is x = h - a
⇒ x = 0 - 4 ⇒ x = -4
x = 4 = 0
(d) Length of latus rectum is 4a = 16.

 Passing through (5, -2) length of the transverse axis is a long x-axis and of length 8 units.
2a = 8 ⇒ a = 4
Since the transverse axis is along x-axis, centre is (0, 0)
Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
Since (5, -2)passes through the parabola,
\(\frac { 25 }{ 16 } -\frac { 4 }{ { b }^{ 2 } } \Rightarrow \frac { 4 }{ { b }^{ 2 } } =\frac { 25 }{ 16 } -1=\frac { 25-16 }{ 16 } =\frac { 9 }{ 16 } \)
∴ \({ b }^{ 2 }=\frac { 16\times 4 }{ 9 } =\frac { 64 }{ 9 } \)
∴ Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ \frac { 64 }{ 9 } } =1\Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 64 } =1\)

End points of latus rectum are (4, -8) and (4, 8)
Focus is the mid-point of (4, -8) and (4, 8)

∴ Focus = \(\left( \frac { 4+4 }{ 2 } ,\frac { -8+8 }{ 2 } \right) \) = (4, 0)
∴ a = 4 and vertex is (0, 0)
Equation of the parabola is y² = 4ax
⇒ y² = 4(4)x
⇒ y² = 16x

SS′ = 2c and 2c = 4; A'A = 2a = 6
c = 2 and a = 3,
b² = a²−c² = 9−4 = 5.
Major axis is along x-axis, since a > b.
Centre (0, 0) and Foci are (±2, 0) 
Therefore, equation of the ellipse is \(\frac { { x }^{ 2 } }{ 9 } \frac { { y }^{ 2 } }{ 5 } =1\)
 

Since axis is parallel to y-axis the required equation of the parabola is (x +1)² = 4a(y+2).
Since this passes through (3, 6) (3 +1)² = 4a(6+ 2) 
a = \(\frac { { 1 } }{ { 2 } } \)
Then the equation of parabola is (x+1)² = 2(y+2) which on simplifying yields,
x²+2x−2y − 3 = 0.

Given vertex A(5, -2) and focus S(2, -2) and the focal distance
AS = a = 3  
Parabola is open left and symmetric about the line parallel to x -axis.
Then, the equation of the required parabola is
(y + 2)² = −4(3)(x − 5)
y² + 4y + 4 = −12x + 60
y² + 4y +12x − 56 = 0  

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

Equation of the parabola is y² = 4ax
Latus rectum LL′ passes through the focus (a, 0)  
Hence the point L is (a, y₁) 
Therefore y₁² = 4a²
Hence y₁ = ±2a
The end points of latus rectum are (a, 2a) and (a, -2a)  

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

Area of the circle = 9π sq.units

πr² = 9π ⇒ r² ⇒ 9 ⇒ r ⇒ 3
Diameters are x + y = 5 ................(1)
and x - y = 1 .................(2)
We know that centre is the point of intersection of diameters.
∴ To find the centre, solve (1) and (2).

⇒ 2x = 6
⇒ x = 3
∴ (1) ⇒ 3 + y = 5
⇒ y = 5 - 3 = 2
∴ Centre is (3, 2)
Hence, equation of the circle is
(x - h)² + (y - k)² = r2
⇒ (x - 3)² + (y - 2)² = 32
\(\Rightarrow x^{2}-6 x+9+y^{2}-4 y+4=9\)
⇒ x² + y² − 6x − 4y + 4 = 0

Given centre is (2, 3)
Let us solve 3x- 2y = 1  .....(1)
and 4x+ y = 27 .....(2)

⇒ x = 5
∴ 3(5) - 2y = 1
⇒ 15 - 2y = 1
⇒ 15-1 = 1
⇒ 14 = 2y
⇒ y = 7
 The circle passes through (5, 7)
[∵ distance between (5, 7) and (2, 3)]
r = \(\sqrt { { (5-2) }^{ 2 }+({ 7-3) }^{ 2 } } \)
= \(\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \)
= \(\sqrt { 9+16 } =\sqrt { 25 } =5\)
Equation of the circle is
(x-h)²+(y-k)² = r²
(x - 2)² + (y - 3)² = 5²
x² - 4x + 4 + y² - 6y + 9 = 25
x² + y² - 4x - 6y + 13 - 25 = 0
x²+y²− 4x − 6y −12 = 0

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

Given r = 5 cm

Since the circle touches the x axis, its centre is (0, ±5)
Equation of the circle is (x - h)² + (y - k)² = r²
⇒ (x - 0)² + (y ± 5)² = 25
\(\Rightarrow x^{2}+y^{2}+\not 25 \pm 10 y=\not 25\)
⇒ x²+y²+10y = 0

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

Coefficient of x² = Coefficient of y² (characteristic (ii) for a second degree equation to represent a circle).
That is, 3 = a + 1 and a = 2 .
Therefore the equation of the circle is 
3x² + 3y² + 6x − 9y + 6 = 0
x² + y² + 2x − 3y + 2 = 0
So, centre is \(\left( -1,\frac { 3 }{ 2 } \right) \) and radius  r =\(\sqrt { 1+\frac { 9 }{ 4 } -2 } \)
=\(\sqrt { \frac { 5 }{ 2 } } \)

As the circle touches both the axes, the distance of the centre from both the axes is 3 units, centre can be (±3, ±3) and hence there are four circles with radius 3, and the required equations of the four circles are
x² + y² ± 6x ± 6y + 9 = 0.

Writing the line 3x+4y = 12, in intercept form yields \(\frac{x}{4}+\frac{y}{3}=1\). Hence the points A and B are (4, 0) and (0, 3) .
Equation of the circle in diameter form is
(x-x₁) (x-x₂)+(y-y₁) (y-y₂) = 0
(x-4) (x-0)+(y-0) (y-3) = 0
x²+y²−4x−3y = 0 .

Equation of the circle passing through the points of intersection of the chord and circle by
Theorem  is x² + y²−16+\(\lambda \)(3x + y + 5) = 0 .
The chord 3x + y + 5 = 0 is a diameter of this circle if the centre\(\left( \frac { -3\lambda }{ 2 } \frac { -\lambda }{ 2 } \right) \) lies on the chord.
So we have 3\(\left( \frac { -3\lambda }{ 2 } \right) \)-\(\frac { -\lambda }{ 2 } \)+5 = 0,
\(\frac { -9\lambda }{ 2 } \)-\(\frac { \lambda }{ 2 } \)+5 = 0,
−5λ + 5 = 0 ,
λ = 1.
Therefore, the equation of the required circle is x² + y²+3x + y −11 = 0.

Let V₁ be the vertex of the parabola and
V₂ be the vertex of the hyperbola.
\(\overset { \_ \_ \_ \_ \_ \_ }{ { F }_{ 1 }{ F }_{ 2 } } \) = 14−2 = 12m, 2c = 12, c = 6
The distance of centre to the vertex of the hyperbola is a = 6−1 = 5
b² = c² - a²
= 36−25 = 11.
Therefore the equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 11 } =1\)

For the parabola,
x²- 4x - 5y -1 = 0
x²- 4x = 5y +1
x²−4x +4 = 5y +1+ 4
(x − 2)² = 5(y +1) which is in standard form.
Therefore 4a = 5 and the vertex is (2, -1) , and focus is \(\left( 2,\frac { 1 }{ 4 } \right) \)
Equation of directrix is 
y-k+ a = 0
y+1+\(\frac { 5 }{ 4 } \)
4y +9 = 0 
Length of latus rectum is 5 units.

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.