the slope of the radius to (4,1) is ... (1 - 0) / (4 - 0) = 1/4
the slope of m (⊥ to the radius) is ... -1 / (1/4) = -4
point-slope form of m is ... y - 1 = -4 (x - 4)
slope-intercept form is ... y = -4 x + 17
Line m is tangent to a circle at the point 4,1 If the circle is centered at the origin, what is the y intercept of the line m?
4 answers
x^2+y^2 = r^2
4^2 + 1^2 = r^2
r^2 = 17
so
x^2 +y^2 = 17^2
what is slope of that circle at (4,1)
2 x dx + 2 y dy = 0
dy/dx = - x/y = -4/1 = -4
that is m in
y = mx+b
y = -4x+b
put in (4,1)
1 = -4(4) + b
b = 17
so the line is
y = -4x + 17
when x = 0, y = 17
4^2 + 1^2 = r^2
r^2 = 17
so
x^2 +y^2 = 17^2
what is slope of that circle at (4,1)
2 x dx + 2 y dy = 0
dy/dx = - x/y = -4/1 = -4
that is m in
y = mx+b
y = -4x+b
put in (4,1)
1 = -4(4) + b
b = 17
so the line is
y = -4x + 17
when x = 0, y = 17
Scott's way was smarter.
thank you , Damon