eqn of line tangent to x^2 + y^2 = 20 at point (-2, 4)

[caseygaspar]

Okay so Im supposed to find the equation in the form of y=mx+b...

The tangent of the circle is x^2+y^2= 20 at the point (-2,4)

I know that i am going to use the negative reciprocal of the slope...but i am a little confused
I plugged in the point into the problem but...
Please help... Thanks guys!

 

[stapel]

The tangent of the circle is x^2+y^2= 20 at the point (-2,4)

To do this, you have to use the definition of "tangent": the tangent line to the circle just touches the circle, but doesn't cross it ("cut into" it) at all. So how must that tangent line and the radius to that point be related? They're perpendicular!

And that's the trick: Find the slope of the line containing the radius, being the line through the center (0, 0) and the point (-2, 4) of tangency; find the perpendicular slope; and then plug that slope and the given point into the point-slope equation. Solve for "y=" to get the slope-intercept form.

Have fun!

Eliz.

 

[caseygaspar]

Thanks sooo much!! it makes sense now....I don't know why i have so much trouble with this....thanks alot!!