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
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)
2 answers
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)
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)
0 answers