<p>In the xy coordinate system, a circle has a center C with coordinates (6,2.5). This circle has exactly one point in common with the x-axis. If the point (3.5, t)
is also a point on the circle, what is the value of t?</p>
<p>I think it is 2.5, however Im not 100% sure on this.</p>
<p>Yeah, it's right. I mean think about. Where else could it be?</p>
<p>2.5 is correct
Knowing that the equation of a circle is (x-h)^2 + (y-k)^2 =r^2, where (h,k) is the center and r is the radius, you can solve the problem using the circle equation. From the description, if the circle has exactly one point in commonm with x-axis, then it is tangent to the axis. therefore the radius is 2.5.
The center is (6,2.5), plug it in along with the radius, and you get the equation of the circle:
(x-6)^2 + (y-2.5)^2=2.5^2
Then, plug in 3.5 for x and t for y. Solve the equation and you get 2.5 as the answer.</p>
<p>Since this is an SAT question it should have a relatively simple solution.
Here it is:
[quote=liebanasuka]
From the description, if the circle has exactly one point in commonm with x-axis, then it is tangent to the axis. therefore the radius is 2.5.<a href="remember%20-%20the%20center's%20y-coordinate%20is%202.5">/quote</a>.
If you draw a vertical line through x=3.5 it'll be tangent to the circle in
(3.5, 2.5) because x=3.5 is exactly 2.5 away from x=6:
6 - 3.5 = 2.5 = radius.
(3.5, 2.5) is (3.5, t)
t = 2.5</p>
<p>is this from the blue book</p>
<p>It's from the October 2007 PSAT.</p>
<p>Stand corrected. Thanks, Quicksand!</p>