<p>This was on a precalc quiz (teacher is senile)</p>
<p>Find the equation of the tangent line, passing thru the origin, of the circle centered at (2,1) with radius 2.</p>
<p>I probably won't get any replies, but I figured I'd see if anyone posted what they think... my brother is a physics major at ND, just finished his sophmore year, and he said that the question was ridiculous for a Precalc class.</p>
<p>x=0? lol...</p>
<p>I've seen this problem before in Calc I, but never has a vertical tangent come up to be a possible solution...</p>
<p>Whoops... yeah my bad, there are two, but one of them is the Y axes, and you need to find both.</p>
<p>jesus, i thought finding the equation of a tangent line (a.k.a. derivative of the equation) was reserved for calculus. Besides, dualityim, there are actually two equations of tangent lines that pass thru the origin. One has a slope of negative .5, and the other, like you said, is the vertical tangent at x=0. Valikor2, your teacher is not only senile, but he/she is also equivocal.</p>
<p>Yeah my precalc H teacher gave a similar question.</p>
<p>x=0
y=-1/2x</p>
<p>While this is certainly a multi-concept problem, it is not a calculus problem. It is not a precalculus level problem either; it is a difficult geometry problem. Use a right triangle to find the distance between the origin and the unknown point of tangency, (x,y). Thus, x^2+y^2=d^2. Since (x-2)^2+(y-1)^2=2^2, you subtract to find 2x+y=(d^2+1)/2. Combine this with x^2+y^2=d^2, to find x and y. Then, find slope and you are done.</p>
<p>The numbers work out nicely. In any case, this is an involved problem.</p>
<p>Hmm, I saw it done using calc... I'm far too tired to think about it at the moment and see what you folks have posted... maybe tommorow...</p>