I expect you meant to say
(2,-√5)
man, some of the weird notations folks come up with!
The line passing through the origin with slope m is
y = mx
You have a point, so you can figure the slope, right?
The tangent line to a circle at any point (x,y) is -x/y
SO, you want to use the point-slope form for a line through (2,-√5) with slope 2/√5
y+√5 = 2/√5(x-2)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+x^2%2By^2%3D9%2C+y+%3D+2%2F%E2%88%9A5%28x-2%29-%E2%88%9A5
A circle of radius 3 is centered at the origin and passes through the point (2,−5‾‾√).
(a) Find an equation for the line through the origin and the point (2,−5‾‾√).
y=
(b) Find an equation for the tangent line to the circle at (2,−5‾‾√).
y=
1 answer
1 answer