A circle which meets the x axis at exactly one point has a center at x equal to 6 and y to 2.5 --- what is the value of y for the circle when x is 3.5?

In the xy-coordinate 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? The formula of a circle is (x - a)2 + (y - b)2 = r2 Where r is the radius, and the centre is at (a,b). This is all down to Pythagoras' theorum - see if you can see why. Anyway, the question gives us the values of a and b as 6 and 2.5. Since the x axis is tangential to the circle, we can also deduce that r = b. This completes the formula: (x - 6)2 + (y - 2.5)2 = 2.52 = 6.25 To answer the question, substitute in x = 3.5 and y = t so that we can find t. (3.5 - 6)2 + (t - 2.5)2 = 6.25 (-2.5)2 + (t - 2.5)2 = 6.25 6.25 + (t - 2.5)2 = 6.25 (t - 2.5)2 = 0 t - 2.5 = 0 t = 2.5 We reached the answer by a purely algebraic method. A sketch graph and some clear reasoning would have served just as well: clearly (3.5,t) is level with the centre of the circle, because its displacement in the x direction is equal to the radius r.