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

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

The circle has center \(\displaystyle (2,4)\) and radius 5, so substitute:

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

\(\displaystyle (x-2)^{2}+ (y-4)^{2}= 25\)

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:

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

\(\displaystyle \left ( \frac{-3+9}{2}, \frac{7+5}{2} \right )\)

\(\displaystyle (3, 6)\)

Therefore, \(\displaystyle h = 3\) and \(\displaystyle k = 6\)

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using \(\displaystyle (3, 6)\) and \(\displaystyle (9, 5)\). We will concern ourcelves with finding the square of the radius \(\displaystyle r^{2}\):

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

\(\displaystyle r^{2}= \left (9-3\right )^{2} + \left ( 5-6\right )^{2}\)

\(\displaystyle r^{2}=6^{2} + \left (-1\right )^{2}\)

\(\displaystyle r^{2}=36+ 1= 37\)

Substitute: 

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

\(\displaystyle (x-3)^{2}+ (y-6)^{2}=37\)

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

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

The circle has center \(\displaystyle (-3, 2)\) and radius 4, so substitute:

\(\displaystyle (x-(-3))^{2}+ (y-2)^{2}= 4^{2}\)

\(\displaystyle (x+3)^{2}+ (y-2)^{2}= 16\)

\(\displaystyle x^{2}+6x + 9+ y ^{2}-4y+4= 16\)

\(\displaystyle x^{2}+6x + 9+ y ^{2}-4y+4- 16= 16-16\)

\(\displaystyle x^{2}+6x + y ^{2}-4y-3=0\)

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 \(\displaystyle r = 3 \sqrt{2}\).

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

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

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

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