a circle has the equation 3(x-5)^2_3y^2=12. Find the center and radius and graph the circle. find the intercepts, if any.
plz show work
Shouldn't there be a + sign between the 2 and the 3? What you have written is not the equation of a circle.
Yes, my mistake.find the center, radius and graph of circle. then find intercepts
3(x-5)^2+3y^2=12
plz show work
divide everybody by 3
(x-5)^2 + y^2 = 4
centre : (5,0), radius 2
clearly the circle can't reach the y-axis,
(let x= 0, ---> 75 + 3y^2 = 12
3y^2 = - .... , not possible)
for the x-intercepts , let y = 0
3(x-5)^2 = 12
(x-5)^2 = 4
x-5 = ± 2
x = 7 or x = 3
Convert:
TT/3 rad to degrees
To find the center and radius of the given circle equation and graph it, we need to rewrite it in standard form, which is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the center of the circle, and r represents the radius.
Let's start by rewriting the given equation in standard form:
3(x - 5)^2 + 3y^2 = 12
Dividing the entire equation by 12 to simplify:
(x - 5)^2/4 + y^2/4 = 1
Now, we can see that the equation is in standard form. The center of the circle is represented by the values (h, k) in the equation. In this case, (h, k) = (5, 0). Therefore, the center of the circle is C(5, 0).
Next, we need to find the radius of the circle, which is represented by r. It can be determined by taking the square root of the number on the right side of the equation. In this case, r^2 = 4, so r = 2. Hence, the radius of the circle is 2.
Now, let's graph the circle on a coordinate plane:
1. Plot the center point C(5, 0).
2. From the center, move 2 units in all directions (up, down, left, right) to determine the endpoints of the radius.
- Up: (5, 0 + 2) = (5, 2)
- Down: (5, 0 - 2) = (5, -2)
- Left: (5 - 2, 0) = (3, 0)
- Right: (5 + 2, 0) = (7, 0)
3. Connect the endpoints of the radius to form the circle.
The graph of this circle has a center at (5, 0) and a radius of 2 units.
To find the intercepts, we need to substitute y = 0 and x = 0 into the equation:
For the x-intercepts, set y = 0:
3(x - 5)^2 + 3(0)^2 = 12
3(x - 5)^2 = 12
(x - 5)^2 = 4
x - 5 = ±√4
x - 5 = ±2
x = 5 + 2 or x = 5 - 2
x = 7 or x = 3
Therefore, the x-intercepts are (7, 0) and (3, 0).
For the y-intercepts, set x = 0:
3(0 - 5)^2 + 3y^2 = 12
3(-5)^2 + 3y^2 = 12
75 + 3y^2 = 12
3y^2 = 12 - 75
3y^2 = -63 (This means there are no real y-intercepts since we can't take the square root of a negative number.)
Hence, the x-intercepts are (7, 0) and (3, 0), and there are no y-intercepts for this circle.