To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

$\displaystyle 4x+3=5x+44;\, 3=x+44;\, -41=x$

To find the y-coordinate, substitute into one of the equations. Let's use $\displaystyle y_{1}$:

$\displaystyle y=4*-41+3=-164+3=-161$

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at $\displaystyle (x_{0},y_{0})$ is:

$\displaystyle (x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}$

For our data, this means that our equation is:

$\displaystyle (x+41)^{2}+(y+161)^{2}=12^{2}$ or $\displaystyle (x+41)^{2}+(y+161)^{2}=144$

We need to expand this equation to $\displaystyle x^{2}-8x+y^{2}-6y=24$ and then complete the square.

This brings us to $\displaystyle x^{2}-8x+16+y^{2}-6y+9=24+16+9$.

We simplify this to $\displaystyle (x-4)^{2}+(y-3)^{2}=49$.

Thus the radius is 7.

The radius of the circle is equal to the hypotenuse of a right triangle with sides of lengths 5 and 7.

$\displaystyle r^2=5^2+7^2$

$\displaystyle r^2=25+49= 74$

$\displaystyle r=\sqrt{74}$

This radical cannot be reduced further.

Recall that the general form of the equation of a circle centered at the origin is:

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

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

$\displaystyle x^{2}+y^{2}=5^{2}$

$\displaystyle x^{2}+y^{2}=25$

Now, the question asks for the positive y-coordinate when $\displaystyle x=2$.  To solve this, simply plugin for $\displaystyle x$:

$\displaystyle 2^{2}+y^{2}=25$

$\displaystyle 4+y^{2}=25$

$\displaystyle y^{2}=21$

$\displaystyle y=\pm \sqrt{(21)}$

Since our answer will be positive, it must be $\displaystyle \sqrt{(21)}$.

Remember that the general equation for a circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ is $\displaystyle (x-h)^2 + (y-k)^2 = r^2$. 

With that in mind, our original center is at $\displaystyle (\frac{\pi}{2}, \frac{3\pi}{2})$ .

If we move the center $\displaystyle \frac{\pi}{2}$ units to the left, that means that we are subtracting $\displaystyle \frac{\pi}{2}$ from our given coordinates. 

$\displaystyle \frac{\pi}{2} - \frac{\pi}{2} = 0$

$\displaystyle \frac{3\pi}{2} - \frac{\pi}{2} = \frac{2\pi}{2} = \pi$

Therefore, our new center is $\displaystyle (0, \pi)$.

The equation of the circle on the coordinate plane with radius $\displaystyle r$ and center $\displaystyle (h,k )$ is

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

The figure referenced is below:

The center of the circle is at the point of intersection of the diagonals, which, as is the case with any rectangle, bisect each other. Therefore,  looking at the diagonal with endpoints $\displaystyle (0,0)$ and $\displaystyle (9,9)$, we can set $\displaystyle x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the midpoint formula:

$\displaystyle h =\frac{ x_{1}+ x_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

and

$\displaystyle k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

The center of the circumscribing circle is therefore $\displaystyle \left ( \frac{9}{2}, \frac{9}{2} \right )$.

The radius of the circumscribing circle is the distance from this point to any point on the circle. The distance formula can be used here:

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

Since we are actually trying to find $\displaystyle r ^{2}$, we will use the form 

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

Choosing the radius with endpoints $\displaystyle (0,0)$ and $\displaystyle \left ( \frac{9}{2}, \frac{9}{2} \right )$, we set $\displaystyle x_{1} = y_{1} = 0 , x_{2} = y_{2} = \frac{9}{2}$ and substitute:

$\displaystyle r^{2}= \left ( \frac{9}{2} - 0 \right ) ^{2} + \left ( \frac{9}{2} - 0 \right ) ^{2}$

$\displaystyle = \left ( \frac{9}{2} \right )^{2} + \left ( \frac{9}{2} \right )^{2}$

$\displaystyle = \frac{81}{4 } + \frac{81}{4 }$

$\displaystyle = \frac{81}{2}$

Setting $\displaystyle h =k= \frac{9}{2}$ and $\displaystyle r^{2}= \frac{81}{2}$ and substituting in the circle equation:

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

$\displaystyle \left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$, the correct response.

A $\displaystyle 135^{\circ }$ arc of a circle represents $\displaystyle \frac{135}{360 } = \frac{3}{8}$ of the circle, so the length of the arc is three-eighths its circumference. Set up the equation and solve for $\displaystyle C$: 

$\displaystyle \frac{3}{8}C = 6 \pi$

$\displaystyle \frac{8} {3} \cdot \frac{3}{8}C =\frac{8} {3} \cdot 6 \pi$

$\displaystyle C = 16 \pi$

The equation of a circle on the coordinate plane is 

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

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

The radius of a circle can be determined by dividing its circumference by $\displaystyle 2 \pi$, so 

$\displaystyle r = \frac{C}{2 \pi } = \frac{16 \pi }{2 \pi } = 8$

The center of the circle is $\displaystyle (0,0)$, so $\displaystyle h = k = 0$. Substituting 0, 0, and 8  for $\displaystyle h$, $\displaystyle k$, and $\displaystyle r$, respectively, the equation of the circle becomes

$\displaystyle (x- 0)^{2}+ (y - 0)^2 =8^{2}$,

or

$\displaystyle x^{2}+ y^2 =64$.

The unshaded region is a $\displaystyle 45 ^{\circ }$ sector of the circle, making the shaded region a $\displaystyle (360 - 45) ^{\circ } = 305 ^{\circ }$ sector, which represents $\displaystyle \frac{305}{360} = \frac{7}{8}$ of the circle. Therefore, if $\displaystyle A$ is the area of the circle, the area of the sector is $\displaystyle \frac{7}{8}A$. The sector has area $\displaystyle 49 \pi$, so 

$\displaystyle \frac{7}{8}A = 49 \pi$

Solve for $\displaystyle A$:

$\displaystyle \frac{8} {7} \cdot \frac{7}{8}A =\frac{8} {7} \cdot 49 \pi$

$\displaystyle A =56\pi$

The equation of a circle on the coordinate plane is 

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

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

The formula for the area $\displaystyle A$ of a circle, given its radius $\displaystyle r$, is 

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

Set $\displaystyle A = 56 \pi$ and solve for $\displaystyle r^{2}$:

$\displaystyle \pi r ^{2} = 56 \pi$

$\displaystyle \frac{\pi r ^{2} }{\pi}=\frac{ 56 \pi}{\pi}$

$\displaystyle r ^{2} = 56$

The center of the circle is $\displaystyle (0,0)$, so $\displaystyle h = k = 0$. Substituting 0, 0, and 56 for $\displaystyle h$, $\displaystyle k$, and $\displaystyle r^{2}$, respectively, the equation of the circle becomes

$\displaystyle (x- 0)^{2}+ (y - 0)^2 =56$,

or

$\displaystyle x^{2}+ y^2 =56$.

The formula for the equation of a circle is:

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

Where (h,k) is the center of the circle.

$\displaystyle h=0$ and $\displaystyle k=4$

and diameter = 6 therefore radius = 3

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

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

The general equation of a circle is $\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}$, where (h, k) represents the location of the circle's center, and r represents the length of its radius. 

Circle A first has the equation of $\displaystyle (x-4)^{2}+(y+3)^{2}=29$. This means that its center must be located at (4, –3), and its radius is $\displaystyle \sqrt{29}$.

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is $\displaystyle 2\sqrt{29}$.

Now, that we have circle A's new center and radius, we can write its general equation using $\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}$.

$\displaystyle (x-(-2))^{2}+(y-2)^{2}=(2\sqrt{29})^{2}=2^{2}(\sqrt{29})^{2}=4(29)=116$.

$\displaystyle (x+2)^{2}+(y-2)^{2}=116$.

The answer is $\displaystyle (x+2)^{2}+(y-2)^{2}=116$.