Using the procedure of comlpeting the square, write the equation x^2+y^2-6x+8y=0 in the form (x-h)^2+(y-k)^2=r^2.
state the domain and range
where(h,k) is the center of the circle and r is the radius
first,
(x^2-6x + ?)+(y^2+8y+?)=?
we need (-3)^3 in the x, and 4^2 in the y
(x^2-6x+9)+(y^2+8y+16)=9+16
(x-3)^2 + (y+4)^2=5^2
how did u get -3^3 and 4^2?
To write the equation x^2 + y^2 - 6x + 8y = 0 in the form (x - h)^2 + (y - k)^2 = r^2 using the completing the square method, follow these steps:
1. Rearrange the equation by grouping the x and y terms separately:
x^2 - 6x + y^2 + 8y = 0
2. To complete the square for the x terms, take half of the coefficient of x (-6) and square it: (-6/2)^2 = 9. Add this value to both sides of the equation:
x^2 - 6x + 9 + y^2 + 8y = 9
3. To complete the square for the y terms, take half of the coefficient of y (8) and square it: (8/2)^2 = 16. Add this value to both sides of the equation:
x^2 - 6x + 9 + y^2 + 8y + 16 = 9 + 16
4. Simplify the equation:
x^2 - 6x + 9 + y^2 + 8y + 16 = 25
(x^2 - 6x + 9) + (y^2 + 8y + 16) = 25
(x - 3)^2 + (y + 4)^2 = 25
Now, the equation x^2 + y^2 - 6x + 8y = 0 is in the form (x - h)^2 + (y - k)^2 = r^2.
The center of the circle is given by (h, k), which in this case is (3, -4), and the radius is given by r = √25 = 5.
Therefore, the equation x^2 + y^2 - 6x + 8y = 0 represents a circle with center (3, -4) and radius 5.
The domain of the circle is all x-values that make the equation true, which means that x can take any real value.
The range of the circle is all y-values that make the equation true, which means that y can take any real value.