1.X^2 + y^2 +2x+6y=26
what is the center
what is the radius
2.9x^{2} +y^{2} -18x -6y+9=0
what is the center
what are the vertices?
name the two points on the minor axis
3. (x-2)^2/16 - (y-1)^2/4 =1
what are the vertices
please show calculations/steps pls i really need it
I will do the 2nd one, they are all basically the same
9x^2 + y^2 - 18x - 6y + 9 = 0
9(x^2 - 2x + ....) - (y^2 - 6y + ...) = -9
9(x^2 - 2x + 1) + (y^2 - 6y + 9) = -9 + 9 + 9
9(x - 1)^2 + (y - 3)^2 = 9
(x - 1)^2 + (y - 3)^2 / 9 = 1
you have an ellipse with centre (1,3)
a^2 = 1 ----> a =1
b^2 = 9 -----> b = 3
vertices are (2, 3) and (0, 3)
minor axis points: (1, 6) and (1, 0)
graph the equation on
h ttps://www.desmos.com/calculator
to verify. Of course delete the space at the front of the URL
x^2 + y^2 + 2x + 6y = 26
x^2+2x + y^2+6y = 26
x^2+2x+1 + y^2+6y+9 = 26+1+9
(x+1)^2 + (y+3)^2 = 36
center at (-1,-3)
radius = 6
9x^2 + y^2 -18x -6y + 9 = 0
9x^2-18x + y^2-6y = -9
9(x^2-2x+1) + y^2-6y+9 = -9+9(1)+9
9(x-1)^2 + (y-3)^2 = 9
(x-1)^2/1 + (y-3)^2/9 = 1
center at (1,3)
a=1, b=3
vertices at (1,3±3)
(x-2)^2/16 - (y-1)^2/4 =1
center at (2,1)
a=4, b=2
vertices at (2±4,1)
THANK YOU BOTH OF YOU
To find the center and radius of a circle given its equation, we need to convert the equation to a standard form, which is expressed as (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center coordinates and r represents the radius.
Let's solve each equation separately:
1. x^2 + y^2 + 2x + 6y = 26
To complete the squares, we rearrange the equation:
x^2 + 2x + y^2 + 6y = 26
(x^2 + 2x) + (y^2 + 6y) = 26
(x^2 + 2x + 1) - 1 + (y^2 + 6y + 9) - 9 = 26
(x + 1)^2 - 1 + (y + 3)^2 - 9 = 26
(x + 1)^2 + (y + 3)^2 = 36
Comparing this equation to the standard form, we see that the center is (-1, -3) and the radius is the square root of 36, which is 6.
2. 9x^2 + y^2 - 18x - 6y + 9 = 0
Again, let's complete the squares:
9x^2 - 18x + y^2 - 6y = -9
(9x^2 - 18x + 9) - 9 + (y^2 - 6y + 9) - 9 = -9
9(x^2 - 2x + 1) + (y - 3)^2 - 9 = -9
9(x - 1)^2 + (y - 3)^2 = 9
Comparing to the standard form, we find that the center is (1, 3) and the radius is the square root of 9, which is 3.
The equation given does not represent a circle, but rather an ellipse. For ellipses, we find the center, vertices, and other points on the major and minor axes.
3. (x - 2)^2/16 - (y - 1)^2/4 = 1
To get to standard form, we rearrange the equation:
(x - 2)^2/16 - (y - 1)^2/4 = 1
((x - 2)^2/16) - ((y - 1)^2/4) = 1
((x - 2)^2/4^2) - ((y - 1)^2/2^2) = 1
Comparing to (x - h)^2/a^2 + (y - k)^2/b^2 = 1, we see that the center is (2, 1).
To find the vertices, we need to compare the denominators with the major and minor axis formulas.
The major axis is parallel to the x-axis, so a^2 = 4^2, and a = 4.
The vertices are located at (h ± a, k), which gives us (2 + 4, 1) = (6, 1) and (2 - 4, 1) = (-2, 1) as the vertices.
Note that this equation represents an ellipse, so there are no points on a "minor axis." Instead, we have co-vertices, which lie on the minor axis perpendicular to the major axis.
The co-vertices are located at (h, k ± b), which gives us (2, 1 + 2) = (2, 3) and (2, 1 - 2) = (2, -1) as the co-vertices.