Since the vertices are (0, ±7), equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
a = 7 and e = \(\frac { 4 }{ 3 } \)
b² = a²(e² - 1) = 49\(\left( \frac { 16 }{ 9 } -1 \right) =49\left( \frac { 16-9 }{ 9 } \right) \)
= \(49\left( \frac { 7 }{ 9 } \right) =\frac { 343 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 49 } -\frac { { x }^{ 2 } }{ \frac { 343 }{ 9 } } =1\) 
⇒ \(\frac { { y }^{ 2 } }{ 49 } -\frac { 9{ x }^{ 2 } }{ 343 } =1\)

Since the foci are one the x-axis, the equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given conjugate axis = \(\frac34\) (transverse axis)
⇒ 2b = \(\frac34\) (2a) ⇒ b = \(\frac{3a}4\) ⇒ b² = \(\frac { { 9a }^{ 2 } }{ 16 } \)
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9{ a }^{ 2 } }{ { 16a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
∴ e = \(\frac { 5 }{ 4 } \)

Given equation of tangent is y = 3x + 1
The condition for any line y = mx + c to be a tangent to y² = 4ax is c = \(\frac{a}{m}\)
1= \(\frac{a}{m}\) ⇒ 1 = \(\frac{a}{m}\) ⇒ a = 3
Length of the latus rectum is 4a = 4(3) = 12 units.

Given circle is x² + y² - 8x - 9y + 12 = 0
Length of the tangent = \(\sqrt { { 2 }^{ 2 }+({ -3) }^{ 2 }-8(2)-9(-3)+12 } \)
= \(\sqrt { 4+9-16+27+12 } \)
= \(\sqrt { 36 } \) = 6 unit

Given circle is  x² +y² + 2x - 3y - 8 = 0
Equation of tangent is xx₁ + yy₁ + 1(x + x₁) -\(\frac{3}{2}\)
(y + y₁) - 8 = 0
AE(2, 3), the tangent is
x(2) + y(3) + x + 2 - \(\frac{3}{2}\) (y + 3) -8 = 0
⇒ 3x + 3y + 2 - \(\frac{3y}{2}\) - \(\frac92\) - 8 = 0
Multiply by 2 we get,
⇒ 6x + 6y + 4 - 3y - 9 - 16 = 0
⇒ 6x + 3y - 21 = 0

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 = 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 = 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 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.

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

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