Find the center (h,k) and the radius (r) of the circle with the given equations:
1. 2(x-2)^2+2(y+5)^2=28
2. 4x^2+4y^2-12x+16y-5=0
#1 is easy. just read them out of the equation. Divide by 2 to make it easier.
#2: (4x^2 - 12x) + (4y^2 + 16y) = 5
or
4(x^2 - 3x) + 4(y^2 + 4y) = 5
(x^2 - 3x) + (y^2 - 4y) = 5/4
Now, you need to make the parenthesized terms perfect squares by adding extra constants on both sides of the equation:
(x^2 - 3x + 9/4) + (y^2 - 4y + 4) = 5/4 + 9/4 + 4
(x - 3/2)^2 + (y - 2)^2 = 15/2
now you can read off h,k r^2 from that
8,9), x + 2y = 7 please solve this for me
To find the center (h,k) and the radius (r) of a circle given its equation, we need to rewrite the equation in the standard form of a circle equation: (x - h)^2 + (y - k)^2 = r^2.
Let's find the center and radius for each given equation:
1. 2(x-2)^2 + 2(y+5)^2 = 28
Start by dividing both sides of the equation by 2:
(x - 2)^2 + (y + 5)^2 = 14
Now, we can see that the equation is in the form (x - h)^2 + (y - k)^2 = r^2.
Comparing the equation to the standard form, we have:
h = 2, k = -5, and r^2 = 14
The center of the circle is (h, k) = (2, -5), and the radius (r) is given by r = √(14).
2. 4x^2 + 4y^2 - 12x + 16y - 5 = 0
To rewrite the equation in standard form, let's complete the square separately for the x and y terms:
4x^2 - 12x + 4y^2 + 16y - 5 = 0
Group the x terms and the y terms:
(4x^2 - 12x) + (4y^2 + 16y) - 5 = 0
Now, we need to complete the square for each group:
4(x^2 - 3x) + 4(y^2 + 4y) - 5 = 0
To complete the square, we take half of the coefficient of the x or y term, square it, and add it inside the parentheses. Also, since we added something inside the parentheses, we need to subtract it on the other side to maintain balance:
4(x^2 - 3x + (-(3/2))^2) + 4(y^2 + 4y + (2)^2) - 5 - 4(-(3/2))^2 - 4(2)^2 = 0
Simplifying further:
4(x^2 - 3x + (9/4)) + 4(y^2 + 4y + 4) - 5 - (9/2) - 16 = 0
Group the perfect square terms and simplify the constants:
4(x^2 - 3x + (9/4)) + 4(y^2 + 4y + 4) - 5 - (9/2) - 16 = 0
4(x - (3/2))^2 + 4(y + 2)^2 - (13/2) - 16 = 0
4(x - (3/2))^2 + 4(y + 2)^2 - 45/2 = 0
Now, the equation is in standard form as (x - h)^2 + (y - k)^2 = r^2.
Comparing the equation to the standard form, we have:
h = 3/2, k = -2, and r^2 = 45/2
The center of the circle is (h, k) = (3/2, -2), and the radius (r) is given by r = √(45/2).