The standard form of the equation of the circle with center at $\displaystyle (h, k)$ and radius $\displaystyle r$ is 

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

Set $\displaystyle h = 7, k = - 4 , r =9$: the correct equation is

$\displaystyle ( x-7) ^{2}+ [y-(-4)]^{2} = 9 ^{2}$

or 

$\displaystyle (x-7) ^{2}+ (y+4)^{2} = 81$

Rewrite the equation of the circle in the form

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

The radius of the circle will be $\displaystyle r$.

Move the constant to the right by subtracting 9 from both sides; also, reorganize the variable terms at left so as to place $\displaystyle x$- and $\displaystyle y$-terms together:

$\displaystyle x^{2}+y^{2}+6x-12y+9=0$

$\displaystyle x^{2}+y^{2}+6x-12y+9 - 9 =0 - 9$

$\displaystyle x^{2}+6x +y^{2}-12y = -9$

Next, complete the squares. First, we will put blanks after the second and fourth terms (grouping for the sake of readability:

$\displaystyle \left (x^{2}+6x+ \underline{\; \; \; \; \; \; \; } \right )+\left (y^{2}-12y + \underline{\; \; \; \; \; \; \; } \right )= -9$

The blanks will be filled with numbers that will complete two perfect square trinomials. In the first blank will be the square of half of 6, which is

$\displaystyle \left ( \frac{6}{2} \right )^{2} = 3^{2} = 9$;

in the second blank, the square of half of $\displaystyle -12$, which is 

$\displaystyle \left ( \frac{-12}{2} \right )^{2} = 6^{2} = 36$

Add these numbers to both sides:

$\displaystyle (x^{2}+6x+9)+(y^{2}-12y + 36) = -9 + 9 + 36$

$\displaystyle (x^{2}+6x+9)+(y^{2}-12y + 36)= 36$

By construction, the trinomials are perfect squares of binomials, and the expression can be rewritten as:

$\displaystyle (x +3)^{2}+(y-6) ^{2}= 36$

or, since the positive square root of 36 is 6,

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

Therefore, $\displaystyle r= 6$, the radius of the circle.

The general form of a circle with its center at point $\displaystyle (h,k)$ and with radius $\displaystyle r$ is 

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

Examine the diagram below:

The center of the circle is $\displaystyle (6,6)$ and the radius is 6, so set $\displaystyle h =k= r = 6$ in the circle equation:

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

or

$\displaystyle (x-6)^{2}+ (y-6)^{2} = 36$

The general form of a circle with its center at point $\displaystyle (h,k)$ and with radius $\displaystyle r$ is 

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

The segment with endpoints at the origin and $\displaystyle (0, -9)$ is a diameter of the circle, so the circle has diameter 9, and its radius is half this, or $\displaystyle -\frac{9}{2}$. The center of the circle is the midpoint of this segment, which is at $\displaystyle \left ( 0, -\frac{9}{2} \right )$.

The diagram showing the center is below.

Set $\displaystyle h = 0 , k = -\frac{9}{2}, r = \frac{9}{2}$. The equation is

$\displaystyle \left [x-0 \right ]^{2}+\left[ y-\left ( - \frac{9}{2} \right ) \right]^{2} = \left ( \frac{9}{2} \right )^{2}$

or

.$\displaystyle x^{2}+ \left (y+ \frac{9}{2} \right )^{2} = \frac{81}{4}$

Rewrite the equation of the circle in the form

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

The coordinates of the center of the circle will be $\displaystyle (h, k)$.

Move the constant to the right by subtracting 9 from both sides; also, reorganize the variable terms at left so as to place $\displaystyle x$- and $\displaystyle y$-terms together:

$\displaystyle x^{2}+y^{2}+6x-12y+9=0$

$\displaystyle x^{2}+y^{2}+6x-12y+9 - 9 =0 - 9$

$\displaystyle x^{2}+6x +y^{2}-12y = -9$

Next, complete the squares. First, we will put blanks after the second and fourth terms (grouping for the sake of readability:

$\displaystyle \left (x^{2}+6x+ \underline{\; \; \; \; \; \; \; } \right )+\left (y^{2}-12y + \underline{\; \; \; \; \; \; \; } \right )= -9$

The blanks will be filled with numbers that will complete two perfect square trinomials. In the first blank will be the square of half of 6, which is

$\displaystyle \left ( \frac{6}{2} \right )^{2} = 3^{2} = 9$;

in the second blank, the square of half of $\displaystyle -12$, which is 

$\displaystyle \left ( \frac{-12}{2} \right )^{2} = 6^{2} = 36$

Add these numbers to both sides:

$\displaystyle (x^{2}+6x+9)+(y^{2}-12y + 36) = -9 + 9 + 36$

$\displaystyle (x^{2}+6x+9)+(y^{2}-12y + 36)= 36$

By construction, the trinomials are perfect squares of binomials, and the expression can be rewritten as:

$\displaystyle (x +3)^{2}+(y-6) ^{2}= 36$

or, since the positive square root of 36 is 6,

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

The center of the circle is located at the point $\displaystyle (-3, 6)$.

First, write the standard form of the equation of the circle, which, if it has center at $\displaystyle (h, k)$ and radius $\displaystyle r$, is 

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

Set $\displaystyle h= -3,k= 8, r =7$ - the equation is

$\displaystyle [(x- (-3)]^{2}+( y- 8 )^{2} = 7^{2}$

or

$\displaystyle (x+3)^{2}+ ( y-8 ) ^{2} = 7^{2}$

The general form of the equation of a circle is

$\displaystyle x^{2}+ y^{2}+ Ax + By+ C = 0$ 

for some real values of $\displaystyle A,B,C$.

Expand the squares of the binomials according to the binomial square pattern:

$\displaystyle (x^{2} +2 \cdot x \cdot 3 +3^{2})+(y^{2}- 2 \cdot y \cdot 8 +8 ^{2} )= 7^{2}$

$\displaystyle (x^{2}+ 6x +9 )+(y^{2}-16y +64 )=49$

Rearrange and collect like terms:

$\displaystyle x^{2}+ 6x +9 + y^{2}-16y +64 =49$

$\displaystyle x^{2}+ y^{2} +6x - 16y+9 +64 =49$

$\displaystyle x^{2}+ y^{2}+ 6x - 16y+73 =49$

Subtract 49 from both sides:

$\displaystyle x^{2}+ y^{2}+ 6x - 16y+73 -49 =49 - 49$

$\displaystyle x^{2}+ y^{2}+ 6x - 16y +24 =0$,

the general form of the equation.

The general form of a circle with its center at the origin is

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

The easiest way to determine the value of $\displaystyle r^{2}$, given that the point $\displaystyle (4, 9 )$ is on the graph, is simply to substitute the values:

$\displaystyle 4^{2}+ 9^{2} = r^{2}$

$\displaystyle 16+81 = r^{2}$

$\displaystyle 97= r^{2}$

The equation of the circle is

$\displaystyle x^{2}+ y^{2} = 97$.