# The small circle has equation x^(2)+y^(2)-2x-4y-4=0. The great circle equation have the same radius with the diameter of the small circle is ...?

Feb 13, 2018

If I understand your question correctly, the large circle has a radius of 6. If the two circles share the same center, the equation of the large circle is ${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 36$

#### Explanation:

Get the equation of the circle from general form to standard form by completing the two squares:

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

Rearrange terms, move constant to other side of the equation:

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

Complete the square for the $x$ terms and the $y$ terms:

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

Simplify

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

This is the standard form of the circle. ${r}^{2} = 9$ so $r = 3$. If the radius of the small circle is three, then the diameter of the small circle is 6, which in your problem, is the radius of the "great circle". If you need the equation of the large circle, and they share the same center, then the equation is:

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = {r}^{2} = {6}^{2} = 36$

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 36$