# How do you find the center, radius, x- and y- intercepts and graph of the circle : x^2+y^2-2x-4y-4=0?

Jul 9, 2015

You convert the equation to standard form and use the values of $h$ and $k$ to calculate these values.

#### Explanation:

Step 1. Convert the equation to standard form.

The standard form for the equation is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$.

We make the conversion by "completing the square".

${x}^{2} + {y}^{2} - 2 x - 4 y - 4 = 0$

${x}^{2} + {y}^{2} - 2 x - 4 y = 4$

$\left({x}^{2} - 2 x\right) + \left({y}^{2} - 4 y\right) = 4$

$\left({x}^{2} - 2 x + 1\right) - 1 + \left({y}^{2} - 4 y + 4\right) - 4 = 4$

${\left(x - 1\right)}^{2} - 1 + {\left(y - 2\right)}^{2} - 4 = 4$

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 4 + 1 + 4$

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 9$

Step 2. Locate the centre.

The centre is at ($h , k$) = ($x , y$) = ($1 , 2$).

The radius is $r = \sqrt{9} = 3$.

Step 4. Calculate the $y$-intercepts.

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 9$

Set $x = 0$.

${\left(0 - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 9$

$1 + {\left(y - 2\right)}^{2} = 9$

${\left(y - 2\right)}^{2} = 9 - 1 = 8$

y-2 = ±sqrt8 = ±2sqrt2

y = 2±2sqrt2

The $y$-intercepts are at ($0 , 2 - \sqrt{2}$) and ($0 , 2 + \sqrt{2}$).

Step 4. Calculate the $x$-intercepts.

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 9$

Set $y = 0$.

${\left(x - 1\right)}^{2} + {\left(0 - 2\right)}^{2} = 9$

${\left(x - 1\right)}^{2} + 4 = 9$

${\left(x - 1\right)}^{2} = 9 - 4$

${\left(x - 1\right)}^{2} = 5$

x-1 = ±sqrt5

x = 1 ±sqrt5

The x-intercepts are at ($1 - \sqrt{5} , 0$) and ($1 + \sqrt{5} , 0$).