<p>Hey guys I found this question extremly hard:
In the xy- cordinate system, a circle has 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>It is a student produced response question and 2.5 was the answer. Why?</p>
<p>K it helps to know the equation of a circle for this equation</p>
<p>equation of a circle is (x-h)^2 + (y-k)^2= r^2 give that (h, k) is the center</p>
<p>The key to the question is that it states that only 1 point on the circumference is common with the axes which indicates that the circumference extends only so far to brush along the X axis (I suppose you draw an x-y coordinate plane to view this labelling the centre O). hence we know the radius</p>
<p>The radius is = 2.5</p>
<p>Thus rewriting the equation of a circle</p>
<p>(x-6)^2 + (y-2.5)^2 = 2.5^</p>
<p>We have the point (3.5, t) hence we input this into the equation</p>
<p>(3.5-6.5)^2 + (t-2.5)^2 = 6.25</p>
<p>Equation proves to be 2.5.</p>
<p>BE CAREFUL WHEN WORKING WITH THESE!!!</p>
<p>wow thanks so much!</p>
<p>It’s really just using the distance formula twice, one to find the radius, and the other to find t. The equation of the circle makes it easier to understand conceptually though.</p>
<p>no prob…If you come across any more difficult ones I’d be happy to be help (If I can 
)</p>
<p>Equations for conic sections are unnecessary and will probably do more harm than good on the test. Here’s a better way of doing this.</p>
<p>Since the circle has only one point in common with the x axis, that point must be directly below the center, at (6, 0). Anywhere else and the circle would have two points in common with the x axis. We now know the radius is 2.5 - 0 = 2.5 and can apply the distance formula to find t. We know the distance between (3.5, t) and (6, 2.5) is the radius, 2.5, so…</p>
<p>2.5 = sqrt[(6 - 3.5)^2 + (2.5 - t)^2]
6.25 = 2.5^2 + (2.5 - t)^2
0 = (2.5 - t)^2
t = 2.5</p>
<p>well thats true. I got 18.5 the 1st time around but i guess I was working way too fast. </p>
<p>The distance line formula is a better pick tho I must say.</p>
<p>Faster way: the value of the distance formula.</p>
<p>If it shares one point in common with the x axis, you know it touches it ONCE, which means the xaxis is TANGENT to the circle. Thus, the distance is the radius, in which case the distance is 2.5.</p>
<p>so 2.5 = rt((6-3.5)^2 + (2.5-t)^2)</p>
<p>solving gives you t = 2.5</p>