<p>I still dont see why it has to be in the top right. If the slope is -1, it can still be anywhere, but itll go through some axis. Does it say that segment doesnt go through any axes?</p>
<p>nbafan - with a slope of negative one the line is always gonna look the same, but it can move around the graph. In order for it to be tangent to the circle, it's got to be in the upper right or the lower left. The only correct coordinants are the ones given for the upper right.</p>
<p>It can be in any of the other places, but those aren't in the answers.</p>
<p>Two reasons why I think this is not a CB question:
1. There are two possible [positive] answers for t.
- The answers don't go in consistent order (ascending or descending).</p>
<p>++++++++++++++++++</p>
<p>Thanks to the courtesy of Yeti Crab:
<a href="http://img.photobucket.com/albums/v634/sushiISgood/quest.jpg%5B/url%5D">http://img.photobucket.com/albums/v634/sushiISgood/quest.jpg</a></p>
<p>For a square with a side equal to "a" its diagonal is equal to (a)sqrt(2).</p>
<p>Mark the origin "Q", point t "T", the center of the circle - "O".</p>
<p>I. For the lower T:
QT = QO - OT
t sqrt(2) = sqrt(2) - 1
t = 1 - 1/sqrt(2)
t = 1 - sqrt(2) / 2</p>
<p>II. For the upper T:
QT = QO + OT
t sqrt(2) = sqrt(2) +1
t = 1 + 1/sqrt(2)
t = 1 + sqrt(2) / 2</p>
<p>Here's how I solved it (took a minute, but I figured it out eventually!):</p>
<p>First, draw the circle.
Draw a line connecting point T(point where the tangent has a slope of -1) to the origin. </p>
<p>Formulas we know: Since AB is tangent to the circle at point T, if you draw a line through the circle that is perpendicular to AB, the slope of that will be 1 (negative reciprocal). Also, in a 45-45-90 degree triangle of hypotenuse h, the side length of the edge is h/(sqrt2).</p>
<p>Plan: Split the line in two. First line is from the center of the circle to the origin. Second line from center to point T. Add the lengths up to find the length of the line from the origin to T. This creates the hypotenuse of a 45-45-90 degree triangle. Divide by sqrt2 to get the x and y coordinates of T.</p>
<p>Now, draw a line connecting the center of the circle to (0,0). Notice that the distance from the radius (1,1) to (0,0) is the sqrt of 2 due to the Pythagorean theorem.</p>
<p>Next step: Distance of center of circle to T: well, that distance is the radius, 1.</p>
<p>So the distance of the line segment from (0,0) to T is 1 + sqrt 2. That is the hypotenuse of a 45-45-90 degree triangle. The legs of the triangle show the x and y coordinates of T. You know it's a 45-45-90 because the slope of the hypotenuse is 1, so x and y are the same. </p>
<p>To find the x and y coordinates, use the Pythagorean theorem to get that x is (1 + sqrt2) / 2 and so is y.</p>
<p>And there's the answer.</p>
<p>Why are people still explaining this problem?</p>
<p>
[quote]
Why are people still explaining this problem?
[/quote]
Because they only look at the first page.</p>
<p>^ lol</p>
<p>Calculus does make this a lot easier (I know its old).</p>
<p>liebenasuka, I actually did read through all the responses, but I was explaining the problem in a way that I hope other people would understand clearly.</p>
<p>bigweight, is this Cb material?
or did u say it was pr? aw this is pretty hard, my poe guess would be B lol.
blah -1</p>
<p>It seems like calculus makes it harder.</p>
<p>Given:
(x-1)^2 + (y-1)^2 = 1 (Assume the circle is in the first quadrant.)
dy/dx = -1
Find x and y.</p>
<p>(x-1)^2 + (y-1)^2 = 1
2(x-1) + 2(y-1)dy/dx = 0
2x-2+(2y-2)dy/dx = 0
dy/dx = -1
2x-2-2y+2 = 0
2x-2y = 0
x - y = 0
x = y</p>
<p>(x-1)^2 + (y-1)^2 = 1
(x-1)^2 + (x-1)^2 = 1
2(x-1)^2 = 1
(x-1)^2 = 1/2
x-1 = sqrt (1/2) = 1/sqrt2 = (sqrt2) / 2</p>
<p>x = 1 + (sqrt2) / 2
y = 1 + (sqrt2) / 2</p>
<p>T = (1 + (sqrt2) / 2 , 1 + (sqrt2) / 2).</p>
<p>^i did exactly the same thing dchow did,but it costed me a lot of time(2min 41 sec).if this kinda math sh0ws up in da real thing,my dream of 800 would b just a mirage.god help me :( .
We could just find out da equation of the straight line that c0nnects the center of da circle and the tangent poing.then just solving da equation w0uld do</p>
<p>This is obviously a Math IIC question.</p>
<p>@amciw - in January 2008 SAT I math was a question of comparable difficulty.
It had a square with four circles "tucked" in each corner and tangent to each other, and a small circle in the center tangent to all four big ones.
I can't divulge the complete question, but you might try mining its discussion in that test's "official" thread.</p>
<p>@gluttony - this is a neat idea.
Since the line is y=x, and
the circe is (x-1)^2 + (y-1)^2 = 1,
2(x-1)^2 = 1
x-1 = +-sqrt(2) / 2
x = 1 +-sqrt(2) / 2
"upper" t = 1 + sqrt(2) / 2</p>
<p>
Something else is funny about this thread: at the moment there are 55 Replies and 56 Views of it.
We shall find a cake and eat one too. :D</p>
<p>
</p>
<p>Lels, I must have been on something when I said that. I think the tangent led me to think it was trigonometry, which is Math IIC.</p>
<p>
[quote]
We shall find a cake and eat one too.
[/quote]
</p>
<p>The cake is a lie.</p>
<p>IT IS NOT MATH II....Its SAT I Math Section, from PR's 11 practice tests, and I recommend this book, because even though PR might be sometimes easier than blue book, this book is of high difficulty.</p>
<p>OK. I think this could easily be on the SAT I test, it only took me like 15-20 seconds to think through it.</p>
<p>^you put the rest of us to shame :o</p>