Problem: Find the equation of the: a) tangent to the circle
with center (-1,2) at the point (3,1) b) perpendicular bisector
bisector of (AB) for A (2,6) and B(5,-2).
I need help.
a) Any line that is tangent to the circle has a slope that is
the negative inverse of the slope of the radius at the tangent
point.
For the radius: m = (y1 - y2)/(x1 - x2)
m = (2 - 1)/(-1 - 3)
m = 1/-4
m = -1/4
The slope of the tangent line is m = 4
The equation is found by using the slope and one point on
the line. We are given that one point: (3, 1)
y = mx + b
b = y - mx
b = 1 - 4(3)
b = 1 - 12
b = -11
Our equation is y = 4x - 11
b) As with the first part, we need the slope of the line we
were given. m = (y1 - y2)/(x1 - x2)
m = (6 - (-2))/(2 - 5)
m = (6 + 2)/(-3)
m = 8/-3
m = -8/3
The slope of any line perpendicular to this line is m = 3/8
The midpoint of the given line is ((x1 + x2)/2, (y1 + y2)/2)
(x1 + x2)/2 = (2 + 5)/2 = 7/2 = 3.5
(y1 + y2)/2) = (6 + (-2))/2 = 4/2 = 2
The midpoint is (3.5, 2)
We find the y-intercept with the formula b = y - mx
b = 2 - (3/8)3.5
b = 2 - 1.3125
b = 0.6875
Our equation is y = 3/8 x + 0.6875