<p>The coordinates (x,y) of each point on the circle above satisfy the equation x^2 + y^2 = 25. Line l is tangent to the circle at point A. If the x-coordinate of point A is -4, what is the slope of l?</p>
<p>Here is the link to the image in the problem: <a href="http://img64.imageshack.us/img64/8015/76125608.jpg%5B/url%5D">http://img64.imageshack.us/img64/8015/76125608.jpg</a></p>
<p>Can someone please explain how to get the answer to this question?</p>
<p>are you allowed to use derivatives? If you are, here is one way to do it -
y = sqrt(25 - x^2) , dy/dx = -x/sqrt(25 - x^2)<br>
If x = -4, slope = dy/dx(x= -4) = 4/sqrt(25-16) = 4/3.</p>
<p>^ too complicated for the SAT. </p>
<p>tanx = slope so</p>
<p>arctan(3/4) = answer.</p>
<p>I kind of lost you after the dy/dx part, Maxwell. I don’t really know what derivatives are either. :/</p>
<p>By the way, I have a better image of the problem here : <a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs;
<p>Also, can you explain your answer more elaborately, olleger? I don’t understand it. Sorry</p>
<p>well there’s a rule that the tan of the angle that the line forms with the x axis is equal to the slope. So you can use a 3,4,5 triagle to get the y coordinant to be 3. So it is 3 high and 4 left. So inverse tan (or arctan) 3/4 (opp/adj) to get the slope.</p>
<p>Erm, there has to be an easier way! Lol.</p>
<p>lolololol there is = p</p>
<p>(5-3)/(0–4) = 2/4 = 1/2</p>
<p>Well, I forgot to mention that the answer is actually 4/3</p>
<p>Correct. The answer should be 4/3. Here is how to do it w/o calculas - </p>
<p>You have a right triangle, I wish I could draw but I will try to explain - the two sides that form the right angles are 5 (the radius) and L (the line that’s tangent to the circle.) then: tan(alpha) is 3/4. (which is NOT the slope). where alpha is the angle between the radius and the horizontal diameter. The slope is the tangent of the complimentary of alpha. so slope is tan (theta) = ctan(alpha) = 1/3/4 = 4/3.</p>
<p>I know it sounds confusing, because I can’t draw the picture. Otherwise it’s easy to show. I’ll try to draw a diagram.</p>
<p>I have a better and easier way w/o calculas - </p>
<p>use the similarity of two triangles. both are right triangles. the small one, the right angle sides are 3, -4, hypotenuse is 5. The big one, the right angle sides are 5, (radius) and x (tangent line), hypotunuse is the horizontal line that passes the diameter of the circle. (I hope you can draw them. ) </p>
<p>Then: 3/x = 4/5, s x = 15/4. Therefore, slope = 5/x = 5/15/4 = 4/3.</p>
<p>oh wow… that was just me being stupid… lol… it doesnt go through the top…</p>
<p>best way I guess is with arctan.</p>
<p>First, we know that y=3 because when we substitute -4 for x, we get y^2=9, so y=3. (0,0), is the other point we use because when a line is tangent to a circle, then a radius drawn to the point of contact is perpendicular to the tangent at that exact point.Then we use y2-y1/x2-x1 to find the slope of the perpendicular line to line l. (3-0)/(0-4)=-3/4. Therefore the slope of l=4/3.</p>
<p>^ is the best method. I don’t even think there is any trigonometry in the SAT Math (there is trig in the subject tests). To solve this question you need to be good in geometry and think in a different perspective. You can find the missing y component of the intersection by plugging in -4 into x^2+y^2=25 and y comes out as 3. If you draw a line from the origin to the point of intersection, you will see that this line is perpendicular to the tangent line. If you find the slope of the line you drew, you can find the slope of the line that’s perpendicular to it.</p>
<p>Is this from college board?</p>
<p>Oblivi0n, that is exactly what I did. It seems like the easiest and most obvious way.</p>
<p>You can easily solve it using Calculus but the SAT math questions don’t require you to know Calculus and most of the takers don’t know high-leveled math. In my opinion, it’s the most logical way to solve it.</p>
<p>i think u have to apply the concept of tangent. in this case, try to draw a line that passes the origin and the point concerned, i.e. A (-4, 3) (see image for urself, then u will see why the y-coordinate is not -3). The slope of this line should be (3-0) / (-4-0) = 3/ -4.
But, this line is perpendicular to line L (by definition, a tangential line is orthogonal to the line that passes the point concerned and the center of circle). This means that the slope of line L (let it be m) is : m x (-3/4) = 1 => m = 4/3. Which is the answer.
This method is an approach that is ideal in tackling SAT questions, since it requires least time and the concept is rather simple (obviously much less time consuming than calculus method or tan method). Hope this helps.</p>
<p>Bassir: Yeah it is.</p>
<p>I think that the way to do this is to find the slope of the radius at that point on the circle. The slope is the change in y over the change in x. The coordinate of that point is (-4,3) based on the equation and info given. The slope would be y/x, or -3/4. The tangent line is perpendicular to the “radius line.” The slope of a perpendicular line is the inverse reciprocal (flip the fraction and multiply by -1) of the slope of the other line. The inverse reciprocal is 4/3, the slope of the tangent line.</p>
