Answer
Center = (1, -2)
Radius = 3 units
Intercepts:
x-axis: $(1+\sqrt{5}, 0)$ and $(1-\sqrt5, 0)$
y-axis: $(0, -2 + 2\sqrt{2})$ and $(0, -2 - 2\sqrt{2})$
Graph on attached image
Work Step by Step
We have,
$x^{2}+y^{2}-2x+4y-4 = 0$
$(x-1)^{2}-1+(y+2)^{2}-4-4=0$
$[\because (x-1)^{2}-1 = x^{2}-2x \space and\space (y+2)^{2}-4 = y^{2}+4y]$
$\implies (x-1)^{2}+(y+2)^{2}=9$
This is of the form, $(x-h)^{2}+(y-k)^{2}=r^2$, where (h,k) is the center and r is the radius.
$\therefore center = (1, -2)\space and\space radius, r = \sqrt{9} = 3$
To find x-intercepts, replace y with 0 in the original equations, and solve for x.
$\implies x^{2}-2x-4 = 0 \implies $This is a quadratic equation, with a = 1, b = -2, c = -4
Using the formula, $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2}$
$x = \frac{2 \pm \sqrt{(-2)^{2}-4(-4)}}{2}$
$\quad = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt5$
To find y-intercepts, replace x with 0 in the original equations, and solve for y.
$\implies y^{2}+4y-4=0 \implies $ This is a quafratic equation, with a = 1, b = 4, c = -4
Using the aforementioned formula,
$x = \frac{-4 \pm \sqrt{4^{2}-4(-4)}}{2}$
$\quad = \frac{-4 \pm \sqrt{32}}{2} = -2 \pm 2\sqrt{2}$
