   SEARCH HOME Math Central Quandaries & Queries  Question from Randy, a student: Hi! I have a question that goes: * Find the equation of the tangent line at coordinates (-1 , 4) on the circle x^2 + y^2 - 4x - 21 = 0 I would like to learn the fastest way to relate any coordinates of a circle to any possible point of tangency. With lots of thanks, Randy Hi Randy.

Here's how I would do it as fast as I can:

Step 1: put the equation of the circle in (mostly) standard form. to do this, I complete the square, ignoring the radius
calculation:

x2 + y2 - 4x
becomes
(x2 - 4x + 4) + y2
= (x - 2)2 + (y - 0)2
So the center of the circle is at (2, 0).

Step 2: find the slope of the tangent line. this is the negative reciprocal of the radius from the circle's center to the
point of tangency, because the tangent and the radius are perpendicular:

m = - (-1 - 2) / (4 - 0) = 3 / 4.

Step 3: plug the slope and point of tangency into the point-slope form of linear equation:

y - 4 = (3 / 4)(x - (-1))
y = 3x / 4 + 19 / 4.

Finished. Is that quick enough?

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.