The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius. 

The diameter is twice the radius:

\(\displaystyle r = \frac{d}{2} = \frac{7}{2}\)

So plugging in the coordinates the circle equations is:

\(\displaystyle (x-3)^2 + (y-4)^2 =(\frac{7}{2})^2 = \frac{49}{4}\)

The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius. 

The diameter is twice the radius:

\(\displaystyle r = \frac{d}{2} = \frac{12}{2} = 6\)

So plugging in the coordinates the circle equations is:

\(\displaystyle (x+2)^2 + (y-4)^2 =6^2 = 36\)

The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius.

So plugging in the coordinates the circle equations is:

\(\displaystyle (h,k)=(2,3)\) and \(\displaystyle r=4\)

\(\displaystyle (x-2)^2 + (y-3)^2 =4^2 = 16\)

The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius. 

\(\displaystyle (h,k)=(5,6);r=2\)

So plugging in the coordinates the circle equations is:

\(\displaystyle (x-5)^2 + (y-6)^2 =2^2 = 4\)

The equation of a circle in standard form is given as

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

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

In order to get the equation

\(\displaystyle x^2+y^2-6x-14y=42\)

in standard form we must complete the square.

First we group together similar variables

\(\displaystyle x^2+y^2-6x-14y=x^2-6x+y^2-14y=42\)

Then we complete the square by

1. finding half of the coefficients of each variable with degree 1
2. squaring those results
3. and then adding those numbers to the equation.

As such,

\(\displaystyle x^2-6x+(\frac{6}{2})^2+y^2-14y+(\frac{14}{2})^2=42+(\frac{6}{2})^2+(\frac{14}{2})^2\)

\(\displaystyle x^2-6x+9+y^2-14y+49=100\)

And because the square of a difference is given as this equation through factoring

\(\displaystyle (a-b)^2=a^2-2ab+b^2\)

we have

\(\displaystyle x^2-6x+9+y^2-14y+49=(x-3)^2+(y-7)^2=100\)

or

\(\displaystyle (x-3)^2+(y-7)^2=10^2\)

First we must express the equation in standard form we can determine what the radius of our circle will be. The standard form for the equation of a circle is given as follows:

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

Where the point (h,k) gives the center of the circle and r is the radius of the circle, which can be easily determined by taking the square root of \(\displaystyle r^2\) once the equation is in standard form. Our first step is to multiply both sides of the equation by 3 to cancel the division by 3 on the left side:

\(\displaystyle 3\left(\frac{(x+5)^2}{3}+\frac{(y-4)^2}{3}\right)=3(3)\)

\(\displaystyle (x+5)^2}+(y-4)^2=9\)

Now we can see that our equation is in standard form, where h=-5 and k=4, which tells us the coordinates of the center of the circle:

\(\displaystyle Center:(-5,4)\)

We can also determine the radius of the circle by taking the square root of \(\displaystyle r^2\):

\(\displaystyle r^2=9\rightarrow \sqrt{r^2}=\sqrt{9}\rightarrow r=3\)

Now that we know the center of the circle is at (-5,4), and that its radius is 3, we can find the points directly above and below the center by adding 3 to its y-coordinate, and then subtracting 3, respectively, giving us:

\(\displaystyle (-5,7)\) and \(\displaystyle (-5,1)\)

Similarly, to find the points directly to the left and to the right of the center, we subtract 3 from its x-coordinate, and then add 3, respectively, giving us:

\(\displaystyle (-8,4)\) and \(\displaystyle (-2,4)\)

We must begin by recalling the general formula for the equation of a circle.

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

Where circle has center of coordinates \(\displaystyle \small (h,k)\) and radius of \(\displaystyle \small r\).

That means that looking at our equation, we can see that the center is \(\displaystyle \small (3,-2)\).

If \(\displaystyle \small r^2=4\), then taking the square roots gives us a radius of 2.

We then look at our possible choices.  Only two are centered at \(\displaystyle \small (3,-2)\).  Of these two, one has a radius of 2 while one has a radius of 4.  We want the former.

Since the circle is centered at \(\displaystyle (0,0)\) we use the  most basic form for the equation for a circle: 

\(\displaystyle x^2+y^2=r^2\).

Given the circle has a radius of \(\displaystyle 5\) marks, which represent \(\displaystyle 3\) units each, the circle has a radius of \(\displaystyle 15\) units.

We then plug in \(\displaystyle 15\) for \(\displaystyle r\): 

\(\displaystyle x^2+y^2=(15)^2\) and simplify: \(\displaystyle x^2+y^2=225\). 

The point \(\displaystyle (3, 0 )\) is the center of the circle - it is not on the circle.

We can test to see if the other points are actually on the circle by plugging in their x and y values into the equation. For example, to verify that

\(\displaystyle (9, \sqrt{13})\) is actually on the circle, we can plug in \(\displaystyle 9\) for x and \(\displaystyle \sqrt{13}\) for y:

\(\displaystyle \sqrt{13} ^ 2 + (9 - 3)^ 2 = 49\)

\(\displaystyle 13 + 6^2 = 49\)

\(\displaystyle 13 + 36 = 49\)

\(\displaystyle 49 = 49\) this is true, so that point is on the circle.

To quickly figure this out, we can plug in 5 for x and -2 for y and see what happens:

\(\displaystyle (5 - 2 ) ^ 2 + (y + 3 )^ 2 =? \enspace 9\)

\(\displaystyle 3^ 2 + 1 ^ 2 =? \enspace 9\)

\(\displaystyle 9 + 1 = 10 > 9\)

Since the value is greater than 9, this point is outside the radius of this circle.