Upon drawing a rough sketch of the coordinates given, we realise that this is a right-angled triangle.

The three side lengths are 6, 8 and 10 units

The inradius of a circle can be calculated using the formula: $rs=Area$, where s is the semi-perimeter of the triangle 

The Area would be $\frac{1}{2}\times\ 6\times\ 8=24$

and the semi-perimeter would be $\frac{10+6+8}{2}=12$

Giving the inradius to be $\frac{24}{12}=2$ units

Therefore, 2 is the correct answer. 

Given equation of circle = $ x^{2} + y^{2} + 4x - 6y - 3 = 0 $

Center of the circle is (-2,3) and radius of the circle = $ \sqrt{ g^2+f^2-c} = \sqrt{ 4+9+3} = 4 $

Let us assume the point of the intersection of the tangents is ( h,k)

The angle made by the line joining (h,k) to the centre makes an angle of 30 degrees with the tangent, and sin(30) will be the ratio of the radius and the distance between the center and (h,k)

=> $ \sin(30) = \frac{4}{\sqrt{( h+2)^2+(k-3)^2}} $

Squaring on both sides:

$ \frac{1}{4} = \frac{16}{(h+2)^2+(k-3)^2} $

=> $ (h+2)^2+(k-3)^2 = 64 $

When x = 6 => h = 6 => $ 64 + (k-3)^2 = 64 $ => k = 3.

=> required point is (6,3)

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC = ($\frac{1-2}{2}, \frac{1+8}{2}$)

Let the coordinates of vertex D be (x, y)

($\frac{x+3}{2}, \frac{y+4}{2}$) = ($\frac{1-2}{2}, \frac{1+8}{2}$)

x = -4 and y = 5

The answer is option D.

Equation of circle $x^2 + y^2 + 2gx + 2fy + c = 0 $

It passes through $ (0,0),(4,0),(3,9). $

$ \Rightarrow \text{Substitute } (0,0):\ c = 0 $

$ \Rightarrow \text{Substitute } (4,0):\ 16 + 0 + 8g + 0 = 0 \ ;\ g = -2 $

$ \Rightarrow \text{Substitute } (3,9):\ 9 + 81 - 12 + 18f = 0 \ ;\ f = -13/3 $

$ \text{Radius of circle } r = \sqrt{g^2 + f^2 - c} $

$ \Rightarrow r^2 = \frac{205}{9} $

Therefore, Area = $\pi r^2 = \frac{205\pi}{9} $

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint of (2,1) and (-3,-4) lie on x + 9y + c = 0

midpoint of (2,1) and (-3,-4) = $\left( \frac{2-3}{2}, \frac{1-4}{2} \right) = \left( -\frac12, -\frac32 \right) $

Keeping this point in the above line equation, we get c = 14

In any right triangle, the circumradius is half of the hypotenuse.

Here,  

$L = \frac{1}{2} \times \text{(length of the hypotenuse)}$

$L = \frac{1}{2}\left(\sqrt{15^{2} + 9^{2}}\right)$

= $\frac{1}{2}\sqrt{306}$

= $\frac{1}{2} \times 17.49$

= $8.74$

Hence, the integer close to $L$ is $9$.