algebra 2 help?

Question

Write the standard equation, the center, and the radius for the circle:
x^2 + y^2 - 4x + 6y + 13 = 4/9

Answer 1

General form for a circle:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is center and r = radius


x^2 + y^2 - 4x + 6y + 13 = 4/9
(x^2 - 4x) + (y^2 + 6y) = (4/9) - 13

Use completing the square for x and y to find what you need to add on.

(x^2 - 4x + 4) + (y^2 + 6y + 9) = (4/9) - 13 + 4 + 9
(x - 2)^2 + (y + 3)^2 = 4/9
(x - 2)^2 + (y + 3)^2 = (2/3)^2

So the center is (2, -3)
the radius is (2/3)

Answer 2

Right now you have this equation in general form. It requires a bit of algebra to make it into standard form.


First group all x's together and all y's together.
x^2 - 4x + y^2 +6y + 13 = 4/9

Move the constant to the other side of the equal sign:
x^2 - 4x + y^2 +6y = -113/9

Divide coefficients of x by 2, then square that number and add it to both sides of the equation. Do the same for y.
(x^2 - 4x + 4) + (y^2 +6y +9) = -113/9 + 4 + 9

Factor:

= (x^2 - 4x + 4) + (y^2 +6y +9) = 4/9
= (x - 2)^2 + (y+3)^2 = 4/9

So in standard form you would have: (x - 2)^2 + (y+3)^2 = 4/9

The center is (2, -3) and the radius is the square root of 4/9, which is 2/3. We know this because the equation of a circle is:

(x - h)^2 + (y-k)^2 = r^2

Where the center is (h, k).
Hope this helps!

Answer 3

The idea is to put the equation in its standard form for a circle which looks like this:

(x-a)² + (y-b)² = (radius)²

where (a) & (b) represent the center coordinates. To accomplish this we have to:

Complete the square twice:

x^2 - 4x + ???
y^2 + 6y + ?##?

for ??? we need a 4
and ?##? we need a 9

so the 13 can be broken up to 4+9 with the 4 going to the x part to give (x-2)² an the 9 goes with the y to yield (y+3)²

(x-2)² + (y+3)² = (2/3)²

center of circle is at (2, -3) and the radius is sqrt(4/9) = 2/3 units.

Answer 4

The standard eqn for a circle looks like (x-a)^2 + (y-b)^2 = r^2
where a & b represents the center and r represents the radius.

Group all the x and y together:
x^2-4x + y^2 +6y + 13 = 4/9
(x-2)^2-4 + (y+3)^2-9 + 13 = 4/9
(x-2)^2 + (y+3)^2 = (2/3)^2

center = (2,-3), radius = 2/3

Answer 5

(x^2 - 4x) + (y^2 +6y) = 4/9 - 13
Complete the square
(x^2 - 4x + 4) + (y^2 + 6y + 9) = 4/9 - 13 + 4 + 9
4/9 - 13 + 13 = 4/9
(x - 2)^2 + (y + 3)^2 = 4/9 = standarad equation
the radius is the square root of the number to the right of the equal sign so the square root of 4/9 = 2/3 r = 2/3

Answer 6

x²-4x + y²+6y = 4/9 -13
x²-4x+4 + y²+6y+9 = 4/9-13+13
(x-2)²+(y+3)² = 2/3²

Answer 7

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