A circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ has equation

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\displaystyle \left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\displaystyle \left ( \frac{2+(-8)}{2}, \frac{7+3}{2} \right )$

$\displaystyle (-3, 5)$

Therefore, $\displaystyle h = -3$ and $\displaystyle k = 5$.

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using $\displaystyle (-3, 5)$ and $\displaystyle (2, 7)$. We will concern ourcelves with finding the square of the radius $\displaystyle r^{2}$:

$\displaystyle r^{2}= \left ( x_{2}-x_{1}\right )^{2} + \left ( y_{2}-y_{1}\right )^{2}$

$\displaystyle r^{2}= \left ( 2- (-3)\right )^{2} + \left ( 7-5\right )^{2}$

$\displaystyle r^{2}= 5^{2} +2^{2}$

$\displaystyle r^{2}=25+ 4 = 29$

Substitute: 

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

$\displaystyle (x-(-3))^{2}+ (y-5)^{2}= 29$

$\displaystyle (x+3)^{2}+ (y-5)^{2}= 29$

Expand:

$\displaystyle x^{2}+6x+9+ y^{2}-10y+25= 29$

$\displaystyle x^{2}+6x + y^{2}-10y+34= 29$

$\displaystyle x^{2}+6x + y^{2}-10y+34- 29 = 0$

$\displaystyle x^{2}+6x + y^{2}-10y+5 = 0$

A circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ has equation

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

The center is $\displaystyle (4, -6)$, so $\displaystyle h = 4, k=-6$.

To find $\displaystyle r$, use the circumference formula:

$\displaystyle 2 \pi r = C$

$\displaystyle 2 \pi r = 20 \pi$

$\displaystyle r = 10$

$\displaystyle r^{2} = 100$

Substitute:

$\displaystyle (x-4)^{2}+ (y-(-6))^{2}=100$

$\displaystyle (x-4)^{2}+ (y+6)^{2}=100$

A circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ has the equation

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

The center is $\displaystyle (2, 5)$, so $\displaystyle h = 2, k = 5$.

The area is $\displaystyle 36 \pi$, so to find $\displaystyle r^{2}$, use the area formula:

$\displaystyle \pi r^{2}= A$

$\displaystyle \pi r^{2}= 36 \pi$

$\displaystyle r^{2}= 36$

The equation of the line is therefore: 

$\displaystyle (x-2)^{2}+ (y-5)^{2}= 36$

The general equation of a circle with center $\displaystyle (a,b)$ and radius $\displaystyle r$ is:

$\displaystyle (x-a)^2+(y-b)^2=r^2$

Now, plug in the values given by the question:

$\displaystyle (x-(-4))^2+(y-5)^2=4^2$

$\displaystyle (x+4)^2+(y-5)^2=16$

Write the formula for the equation of a circle with a given point, $\displaystyle (h,k)$.

$\displaystyle (x-h)^2+(y-k)^2=r^2$

The radius of the circle is half the diameter, or $\displaystyle 2.5$.

Substitute all the values into the formula and simplify.

$\displaystyle (x-2)^2+(y-(-3))^2=(2.5)^2$

$\displaystyle (x-2)^2+(y+3)^2=6.25$

The standard form of the equation of a circle is 

$\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$

where $\displaystyle r$ is the radius of the circle.

We can rewrite the equation we are given, which is in general form, in this standard form as follows:

$\displaystyle x^{2}+ y^{2}+ 4x- 6y + 2 = 0$

$\displaystyle x^{2}+ y^{2}+ 4x- 6y = -2$

$\displaystyle \left (x^{2}+ 4x \right )+\left ( y^{2}- 6y \right )= -2$

Complete the squares. Since $\displaystyle \left ( \frac{4}{2} \right )^{2} = 4$ and $\displaystyle \left ( \frac{-6}{2} \right )^{2} = 9$, we do this as follows:

$\displaystyle \left (x^{2}+ 4x +4 \right )+\left ( y^{2}- 6y +9\right )= -2+4+9$

$\displaystyle (x+2)^{2}+(y-3)^{2} = 11$

$\displaystyle r^{2} = 11$, so $\displaystyle r = \sqrt{11}$, and the circumference of the circle is 

$\displaystyle C = 2 \pi r = 2 \pi \sqrt{11}$

Below is the figure with the circle and square in question:

The center of the inscribed circle coincides with that of the square, which is the point $\displaystyle (3,3)$. Its diameter is the length of a diagonal of the square, which is $\displaystyle \sqrt{2}$ times the sidelength 6 of the square - this is $\displaystyle 6 \sqrt{2}$. Its radius is, consequently, half this, or $\displaystyle 3 \sqrt{2}$. Therefore, in the standard form of the equation, 

$\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$,

substitute $\displaystyle h = k = 3$ and $\displaystyle r = 3 \sqrt{2}$.

$\displaystyle (x-3)^{2}+(y-3)^{2} = \left (3 \sqrt{2} \right ) ^{2}$

$\displaystyle (x-3)^{2}+(y-3)^{2} = 3^{2} \left ( \sqrt{2} \right ) ^{2}$

$\displaystyle (x-3)^{2}+(y-3)^{2} = 9 \cdot 2$

$\displaystyle (x-3)^{2}+(y-3)^{2} = 18$

Below is the figure with the circle and square in question:

The center of the inscribed circle coincides with that of the square, which is the point $\displaystyle (4,4)$. Its diameter is equal to the sidelength of the square, which is 8, so, consequently, its radius is half this, or 4. Therefore, in the standard form of the equation, 

$\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$,

substitute $\displaystyle h = k = 4$ and $\displaystyle r = 4$.

$\displaystyle (x-4)^{2}+(y-4)^{2} = 4^{2}$

$\displaystyle (x-4)^{2}+(y-4)^{2} = 16$

The standard form of the equation of a circle is 

$\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$

where $\displaystyle r$ is the radius of the circle.

We can rewrite the equation we are given, which is in general form, in this standard form as follows:

$\displaystyle x^{2}+ y^{2}+ 4x- 8y -11 = 0$

$\displaystyle x^{2}+ y^{2}+ 4x- 8y =11$

$\displaystyle \left (x^{2}+ 4x \right ) +\left ( y^{2}- 8y \right ) =11$

Complete the squares. Since $\displaystyle \left ( \frac{4}{2} \right )^{2} = 4$ and $\displaystyle \left ( \frac{-8}{2} \right )^{2} = 16$, we do this as follows:

$\displaystyle \left (x^{2}+ 4x + 4 \right ) +\left ( y^{2}- 8y + 16\right ) =11 + 4 + 16$

$\displaystyle (x-2)^{2}+(y-4)^{2} = 31$

$\displaystyle r^{2}= 31$, and the area of the circle is 

$\displaystyle A = \pi r^{2} = 31 \pi$

The standard form of the equation of a circle is 

$\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$,

where the center is $\displaystyle (h,k)$ and the radius is $\displaystyle r$.

The center of the circle is the origin, so $\displaystyle h = k = 0$, and the equation is 

$\displaystyle x^{2}+ y^{2} = r^{2}$

for some $\displaystyle r$.

The area of the circle is $\displaystyle 64 \pi$, so

$\displaystyle A = \pi r^{2} = 50 \pi$

$\displaystyle \frac{ \pi r^{2}}{\pi} = \frac{ 64 \pi}{\pi}$

$\displaystyle r^{2} = 64$

We need go no further; we can substitute to get the equation $\displaystyle x^{2}+ y^{2} = 64$.