Question

1. What are the coordinates of the center of the circle and the length of the radius with the equation x^2 -12x + y^2 + 4y = -31?
a. (6, -2); r = 3
b. (-6, -2); r = 9
c. (12, 4); r = 31
d. (-6, 2); r = 3

2.What are the coordinates of the center of the circle and the length of the radius with the equation x^2 -2x + y^2 + 10y = -1?
a. (1, -5); r = 5
b. (-1, -5); r = 25
c. (-1, 5); r = 5
d. (-2, 10); r = 1

Answer 1

Do you know how to complete the square? You need to transform the equations to be of the form \[(x-h)^2 + (y-k)^2 = r^2\] and then you'll be able to directly read off the coordinates of the center (\(h,k\)) and the length of the radius \(r\)

Answer 2

You can do this by looking at the \(x\) and \(y\) terms (not squared). You'll take half of the coefficient of that term, square it, and add that quantity to both sides. This will allow you to replace that variable on the left hand side with (var + coefficient/2)^2. Repeat with all the variables and compare your equation with mine.

I'll do the first one as an example.

\[x^2-12x+y^2+4y = -31\]
\[(x^2-12x + (-12/2)^2) + (y^2 + 4y + (4/2)^2) = -31 + (-12/2)^2 + (4/2)^2\]Now rewrite the expressions on the left hand side as perfect squares
\[(x+(-12/2))^2 + (y+(4/2))^2 = -31 + 36 + 4\]\[(x-6)^2+(y+2)^2 = 9\]
Comparison with my equation shows that \(h=6,~k=-2,~r=3\)
The circle has its center at (\(6,-2)\) and the radius is \(3\).

Answer 3

Thank you so much! This really helps a lot!