find the equation of the circle through (2,-1), tangent to x+y=1 and having its center on y=-2x.
Geometry: circle with tangent line.
A line that is tangent to a circle forms a right angle with the radius at the tangent point.
The slope of that radius has to be the inverse reciprocal of the slope of the tangent line.
Change the tangent line's equation to slope-intercept form.
x + y = 1
y = -x + 1
The slope of the tangent line is -1, so the slope of the radius at the tangent point is 1.
The y-intercept of the radius can be computed using the co-ordinates of the given point, (2, -1).
b = y - mx
b = -1 - (1)2
b = -1 - 3
b = -4
Now we see the equation for the radius is y = x - 4.
Next, we solve the system of equations y = -2x and y = x - 4 to find the intersection
of those two lines. That point is the center of the circle.
y = -2x
y = x - 4
-2x = x - 4
-3x = -4
x = 4/3
y = x - 4
y = 4/3 - 4
y = 4/3 - 12/3
y = -8/3
The center is at (4/3, -8/3)
The length of the radius is the distance from the center, (4/3 , -8/3), and
the tangent point, (2, -1).
Step by step:
(2 - (4/3))^2 = 0.444444
(-1 - (-8/3))^2 = 2.77777
0.444444 + 2.77777 = 3.22222
sqrt(3.2222) = 1.795
Now for the equation of the circle.
(x - a)^2 + (y - b)^2 = r^2
(x - 4/3)^2 + (y - (-8/3))^2 = 1.795^2
(x - 4/3)^2 + (y + 8/3)^2 = 1.795^2