<p>hello everyone,i need an efficient method to solve this problem.i already know how to solve it with a graphical calculator.it's a SAT l standard que.here it goes:</p>
<p>**A circle with center A has its center at (6,-2) and a radius of 4.which of the following is the equation of a line tangent to the circle with center A?</p>
<p>(A) y=3x+2
(B) y=2x+1
(C) y=-x+5
(D) y=-2
(E) y=-6
**</p>
<p>Er.. I just drew it out real quick and its obvious that (E) is the answer... sorry I don't have a better explanation.</p>
<p>EDIT: let me try.. what I did was, since the center is (6, -2) I just added the 4 radius to find the point directly below the center, on the circle. That was (6,-6). y=-6 goes through that point.. so it's tangent..</p>
<p>Yes, it's rather obvious once you draw the circle that y=-6 is the correct answer. It helps to try out the easy equations (y=-2, y=-6 first before you try to graph the more complicated equations). Method: Draw a picture.</p>
<p>any method that doesn't require drawing a picture?</p>
<p>^ You would need to use calculus, because there are an infinite number of lines that are tangent to a circle. It can't get much more efficient than a picture.</p>
<p>^ You would need to use calculus, because there are an infinite number of lines that are tangent to a circle. It can't get much more efficient than a picture.</p>
<p>Using calculus, you can be given any slope and find the tangent line(s).</p>
<p>(x-6)^2 + (y+2)^2 = 16
0 = 2(x-6) + 2(y+2)y'
0 = x-6+(y+2)y'
6-x = (y+2)dy/dx
[6-x] / (dy/dx) = y+2 (dy/dx does not equal 0)</p>
<p>y = -2 + (6-x)/(dy/dx).</p>
<p>Given this, you can plug in values of dy/dx to find the function. For example, if you want to find the tangent line with slope 0, you would have to go back to 6-x = (y+2)dy/dx and get x=6. Then going back to the original function, you get y = 2 and y=-6.</p>
<p>Don't even draw out the picture before you've tested the two horizontal lines. It's evident that E is the answer in like 10 seconds.</p>
<p>when your giving a center of a circle and a radius and asked to find the tangent, just add or subtract the radius to/from the x or y coordinates. You should get 4 points. The lines that contain those points would have to be either horizontal or vertical.
x=2
x=10
y=2
y=-6
just pick the one offered in the answer choice.
It should just take seconds to do in your head.
most of the time, the answer to these questions are vertical or horizontal lines. CB just wants to know if you understand what a tangent line is.</p>
<p>and i think there would've been a function if it wants tangent lines at other areas, not just vert or hort. maybe diagonal, then u have to find perpendicularity and stuff. but its not gona be that complicated on sat =p</p>
<p>edit: woot! perpendicularity 16 letters =O</p>
<p>i don't know why,i like to use calculus for this problem......i solved it using calculus......btw,very important advices you people have offered,THANKS to all</p>
<p>You can do this on a graphing calculator...? What, by plugging the x and y values into the conic equation for a circle and solving for the tanget? That seems like it would take a while, especially if you don't remember the eq. for a circle off the top of your head.</p>
<p>Calculus would take longer than common sense to solve this.</p>
<p>Not all math problems should be solved algebraic equations. Its good to know how to do it, but as far as this question is concerned, common sense is the way to go.</p>
<p>Einstein couldn't do math for ****.</p>