Step 1: Identify the standard formula.
The equation of a circle with centre $(h,k)$ and radius $r$ is given by the central form: $${(x-h)}^2+{(y-k)}^2=r^2$$

Step 2: Substitute the given values.
Here, $h=0$, $k=2$, and $r=2$. Substituting these into the formula: $${(x-0)}^2+{(y-2)}^2=2^2$$

Step 3: Expand and simplify.
$$x^2+(y^2-4y+4)=4$$ $$x^2+y^2-4y+4-4=0$$

Step 1: Identify the standard formula.
Using the central form of the circle equation: $${(x-h)}^2+{(y-k)}^2=r^2$$

Step 2: Substitute the given values.
Here, $h=-a$, $k=-b$, and $r=\sqrt{a^2-b^2}$. $${(x-(-a\left)\right)}^2+{(y-(-b\left)\right)}^2={\left(\sqrt{a^2-b^2}\right)}^2$$ $${(x+a)}^2+{(y+b)}^2=a^2-b^2$$

Step 3: Expand the squares.
$$(x^2+2ax+a^2)+(y^2+2by+b^2)=a^2-b^2$$ Subtract $a^2$ and add $b^2$ to both sides to simplify: $$x^2+y^2+2ax+2by+b^2+b^2=0$$

Step 1: Determine the vertices of the rectangle.
The sides of the rectangle are defined by the intersection of the horizontal and vertical lines.
The vertices are formed by pairing the x and y values: $$A(-5,-1), B(4,-1), C(4,5), \text{and} D(-5,5)$$

Step 2: Choose a diagonal.
A diagonal connects opposite vertices. Let's use the diagonal connecting $A(-5,-1)$ and $C(4,5)$. Since the circle is drawn with this diagonal as its diameter, A and C represent the extremities of the diameter.

Step 3: Apply the diameter form of the circle equation.
The equation of a circle with diameter endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is: $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$ Substitute $(x_1,y_1)=(-5,-1)$ and $(x_2,y_2)=(4,5)$: $$(x-(-5\left)\right)(x-4)+(y-(-1\left)\right)(y-5)=0$$ $$(x+5)(x-4)+(y+1)(y-5)=0$$

Step 4: Expand the equation.
$$(x^2-4x+5x-20)+(y^2-5y+y-5)=0$$ $$x^2+x-20+y^2-4y-5=0$$ $$x^2+y^2+x-4y-25=0$$

Step 1: Find the centre of the circle.
The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$. By comparing the given equation with the general form, we get: $$2g=-6\Rightarrow g=-3$$ $$2f=5\Rightarrow f=\frac{5}{2}$$ The centre of the circle is $C(-g,-f)$. Therefore, the centre is: $$C(3,-\frac{5}{2})$$

Step 2: Use the midpoint formula.
The centre of the circle is the midpoint of its diameter. Let the unknown end of the diameter be $B(x_2,y_2)$ and the given end be $A(-1,3)$. Using the midpoint formula: $$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(x_c,y_c)$$ Substituting the known values: $$\frac{-1+x_2}{2}=3$$ $$-1+x_2=6\Rightarrow x_2=7$$ And for the y-coordinate: $$\frac{3+y_2}{2}=-\frac{5}{2}$$ $$3+y_2=-5\Rightarrow y_2=-8$$

Step 1: Compare with the general equation of a circle.
The general form is $x^2+y^2+2gx+2fy+c=0$. By comparison: $$2g=8\Rightarrow g=4$$ $$2f=10\Rightarrow f=5$$ $$c=p$$

Step 2: Use the radius formula.
The radius $r$ of a circle is given by: $$r=\sqrt{g^2+f^2-c}$$ Given $r=7$, substitute the known values: $$7=\sqrt{4^2+5^2-p}$$

Step 3: Solve for $p$.
Square both sides to remove the square root: $$7^2=16+25-p$$ $$49=41-p$$ $$p=41-49=-8$$

Step 1: Compare with standard form.
The given equation is $y^2=16x$. This is of the standard form $y^2=4ax$ (parabola opening to the right along the positive x-axis).

Step 2: Find the value of $a$.
$$4a=16\Rightarrow a=4$$

Step 3: Extract the properties.

• Focus: For $y^2=4ax$, the focus is $(a,0)$. So, Focus = $(4,0)$.
• Axis of the parabola: Since the square is on $y$, the axis is the x-axis, whose equation is $y=0$.
• Equation of directrix: The directrix is $x=-a$. So, directrix is $x=-4$ or $x+4=0$.
• Length of latus rectum: The formula is $4a$. So, Latus Rectum = $16$.

Step 1: Rewrite into standard form.
The given equation is $3x^2=8y$. Dividing by 3 yields: $$x^2=\frac{8}{3}y$$ This is of the standard form $x^2=4ay$ (parabola opening upwards along the positive y-axis).

Step 2: Find the value of $a$.
$$4a=\frac{8}{3}\Rightarrow a=\frac{2}{3}$$

Step 3: Extract the properties.

• Focus: For $x^2=4ay$, the focus is $(0,a)$. So, Focus = $(0,\frac{2}{3})$.
• Axis of the parabola: Since the square is on $x$, the axis is the y-axis, whose equation is $x=0$.
• Equation of directrix: The directrix is $y=-a$. So, directrix is $y=-\frac{2}{3}$.
• Length of latus rectum: The formula is $4a$. So, Latus Rectum = $\frac{8}{3}$.

Step 1: Identify the standard form.
The focus is of the form $(a,0)$ where $a=6$. Since the y-coordinate of the focus is 0, the axis of the parabola is the x-axis. Because the focus has a positive x-coordinate, the parabola opens to the right.
The directrix is $x=-a$ which matches $x=-6$.
The standard equation for this type of parabola is $y^2=4ax$.

Step 2: Substitute the value of $a$.
Substitute $a=6$ into the standard equation: $$y^2=4\left(6\right)x$$ $$y^2=24x$$

Step 1: Identify the standard form.
The vertex is at the origin $(0,0)$ and the focus is at $(-2,0)$. Since the focus lies on the negative x-axis, the parabola opens to the left. The standard equation for a parabola opening to the left with vertex at the origin is: $$y^2=-4ax$$ where the focus is at $(-a,0)$.

Step 2: Substitute the value of $a$.
From the focus $(-2,0)$, we have $-a=-2\Rightarrow a=2$. Substitute $a=2$ into the equation: $$y^2=-4\left(2\right)x$$ $$y^2=-8x$$

Step 1: Compare with the general equation of a circle.
The general equation is $x^2+y^2+2gx+2fy+c=0$ where the centre is $(-g,-f)$.
Comparing $x^2+y^2-4x-10y+4=0$ with the general form: $$2g=-4\Rightarrow g=-2$$ $$2f=-10\Rightarrow f=-5$$

Step 2: Find the coordinates of the centre.
The centre is $(-g,-f)$, so: $$\text{Centre}=(-(-2),-(-5\left)\right)=(2,5)$$

Step 1: Check the position of the point.
To find the position of a point $(x_1,y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$, substitute the point into the expression. Let $S=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c$.
If $S<0$, it lies inside.
If $S=0$, it lies on the circle.
If $S>0$, it lies outside.

Step 2: Evaluate for (0, 0).
$$S=0^2+0^2-2\left(0\right)-4\left(0\right)=0$$ Since the result is exactly 0, the point satisfies the equation of the circle.

Step 1: Analyze the parabola.
The equation is $x^2=-8y$. This is a parabola opening downwards with standard form $x^2=-4ay$.
Comparing, we get $4a=8\Rightarrow a=2$.

Step 2: Find equations of directrix and latus rectum.
For $x^2=-4ay$:
The directrix is the horizontal line $y=a\Rightarrow y=2$.
The latus rectum is the horizontal line passing through the focus $(0,-a)$, so its equation is $y=-a\Rightarrow y=-2$.

Step 3: Calculate the distance.
The distance between the parallel horizontal lines $y=2$ and $y=-2$ is: $$|2-(-2\left)\right|=|2+2|=4\text{units}$$

Step 1: Find the centre of the given circle.
Concentric circles share the same centre. For the circle $x^2+y^2-6x+4y-3=0$:
$2g=-6\Rightarrow g=-3$
$2f=4\Rightarrow f=2$
The centre is $(-g,-f)=(3,-2)$.

Step 2: Construct the new circle's equation.
The new circle has centre $(3,-2)$ and radius $r=\sqrt{50}$. Its equation is: $${(x-3)}^2+{(y-(-2\left)\right)}^2={\left(\sqrt{50}\right)}^2$$ $$x^2-6x+9+y^2+4y+4=50$$ $$x^2+y^2-6x+4y+13-50=0$$ $$x^2+y^2-6x+4y-37=0$$

Step 1: Use the given point to find the parameter.
Since the parabola $y^2=4ax$ passes through the point $(3,6)$, substitute $x=3$ and $y=6$ into the equation: $$6^2=4a\left(3\right)$$ $$36=12a$$ $$a=3$$

Step 2: Find the length of the latus rectum.
The length of the latus rectum of the parabola $y^2=4ax$ is given by the formula $4a$.
$$\text{Latus Rectum}=4a=4\left(3\right)=12\text{units}$$