Find circle given the center $\left(-4, 9\right)$, the diameter $10$

The standard form of the equation of a circle is $\left(x - h\right)^{2} + \left(y - k\right)^{2} = r^{2}$, where $\left(h, k\right)$ is the center of the circle and $r$ is the radius.

Thus, $h = -4$, $k = 9$.

Since $d = 2 r$, then $2 r = 10$.

Solving the system $\begin{cases} h = -4 \\ k = 9 \\ 2 r = 10 \end{cases}$, we get that $h = -4$, $k = 9$, $r = 5$ (for steps, see system of equations calculator).

The standard form is $\left(x + 4\right)^{2} + \left(y - 9\right)^{2} = 25$.

The general form can be found by moving everything to the left side and expanding (if needed): $x^{2} + 8 x + y^{2} - 18 y + 72 = 0$.

Radius: $r = 5$.

Area: $A = \pi r^{2} = 25 \pi$.

Both eccentricity and linear eccentricity of a circle equal $0$.

The x-intercepts can be found by setting $y = 0$ in the equation and solving for $x$ (for steps, see intercepts calculator).

Since there are no real solutions, there are no x-intercepts.

The y-intercepts can be found by setting $x = 0$ in the equation and solving for $y$: (for steps, see intercepts calculator).

y-intercepts: $\left(0, 6\right)$, $\left(0, 12\right)$

The domain is $\left[h - r, h + r\right] = \left[-9, 1\right]$.

The range is $\left[k - r, k + r\right] = \left[4, 14\right]$.

Standard form/equation: $\left(x + 4\right)^{2} + \left(y - 9\right)^{2} = 25$A.

General form/equation: $x^{2} + 8 x + y^{2} - 18 y + 72 = 0$A.

Graph: see the graphing calculator.

Center: $\left(-4, 9\right)$A.

Radius: $5$A.

Diameter: $10$A.

Circumference: $10 \pi\approx 31.415926535897932$A.

Area: $25 \pi\approx 78.539816339744831$A.

Eccentricity: $0$A.

Linear eccentricity: $0$A.

x-intercepts: no x-intercepts.

y-intercepts: $\left(0, 6\right)$, $\left(0, 12\right)$A.

Domain: $\left[-9, 1\right]$A.

Range: $\left[4, 14\right]$A.