Find the center radius and equation of a circle in standard form given the following conditions:
1. Tangent to 3x+2y=0 at the point (0,0) and passing through (1,-1) and (6,0)
the center is on the line y = 2/3 x (the radius is perpendicular to 3x+2y=0). So the equation is
(x-h)^2+(y - 2/3 h)^2 = r^2
using the given points, we have
(1-h)^2 + (-1 - 2/3 h)^2 = r^2
(6-h)^2 + (0 - 2/3 h)^2 = r^2
solve for h and r, and our circle is
(x-3)^2 + (y-2)^2 = 13
see
http://www.wolframalpha.com/input/?i=plot+(x-3)%5E2+%2B+(y-2)%5E2+%3D+13,+3x%2B2y+%3D+0,+y%3D0,+x%3D0
equation of that line is y=-3/2 x so
slope = -3/2
eqn of circle
(x-k)^2 + (y-h)^2 = r^2
2(x-k)dx +2(y-h)dy = 0
dy/dx = -(x-k)/(y-h)
at (0,0) dy/dx = k/h = -3/2
say
h= - (2/3 )k
at (0,0)
(x- k)^2 +(y+2k/3)^2 = r^2 but x = y = 0
k^2+4k^2/9 = r^2
13 k^2/9 = r^2
so
(x- k)^2 +(y+2k/3)^2 = 13 k^2/9
goes through (6,0)
(6-k)^2 + 4k^2/9 = 13k^2/9
(6-k)^2 = k^2
36 -12 k = 0
k = 3 whew, something useful center at x = 3
now for h
h = -2k/3 = -2
so
center at (3,-2)
for r^2
r^2 = 13 k^2/9 = 13
so
(x-3)^2 + (y+2)^2 = 13
but it said it went through (1,-1)
does it?
(-2)^2 +1^2 = 5, nope
I think it goes through (-1,1)
To find the center, radius, and equation of a circle in standard form given the conditions mentioned, you can follow these steps:
1. Determine the center of the circle:
Since the circle is tangent to the line 3x + 2y = 0 at the point (0,0), the center of the circle must lie on a line perpendicular to this tangent line and passing through (0,0).
The given line's slope is -3/2, so the perpendicular line's slope will be the negative reciprocal: 2/3.
The equation of the perpendicular line passing through (0,0) can be written in point-slope form:
y - 0 = (2/3)(x - 0)
y = (2/3)x
Now, we have two equations: 3x + 2y = 0 and y = (2/3)x.
We can solve them simultaneously to find the point of intersection, which will be the center of the circle.
Substitute y in the equation 3x + 2y = 0 with (2/3)x:
3x + 2(2/3)x = 0
3x + (4/3)x = 0
(9/3)x + (4/3)x = 0
(13/3)x = 0
x = 0
Plug the obtained x-value into y = (2/3)x:
y = (2/3)(0)
y = 0
Therefore, the center of the circle is (0, 0).
2. Determine the radius of the circle:
We can find the distance between the center (0,0) and one of the given points on the circle, say (1,-1), using the distance formula. The distance formula is:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Plugging in the values, we get:
d = sqrt((1 - 0)^2 + (-1 - 0)^2)
d = sqrt(1^2 + (-1)^2)
d = sqrt(1 + 1)
d = sqrt(2)
Therefore, the radius of the circle is √2.
3. Write the equation of the circle in standard form:
The standard form of a circle equation is: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center coordinates, and r represents the radius.
Substituting the known values, we get:
(x - 0)^2 + (y - 0)^2 = (√2)^2
x^2 + y^2 = 2
Hence, the equation of the circle in standard form is x^2 + y^2 = 2.