<p>here's the link
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<p>The answer is B (1,1). If you plot the points, you will find that there is only one circle which could pass through all of them.</p>
<p>Draw that circle on your paper. To me, it seemed like the center is at (1,1). Then, I tested this by making sure the point (1,1) is equidistant from each point, and sure enough, it was.</p>
<p>The triangle in question is a right triangle with right angle at G (calculate the slope of the legs)</p>
<p>When a right triangle is inscribed in a circle, the hypoteneuse forms a diameter.</p>
<p>The center of the circle must lie at the midpoint of the diameter (1,1).</p>
<ol>
<li>The segment joining the points (-2,1) and (4,1) form a diameter of the circle. The center of a circle is the midpoint of any diameter. The average of -2 and 4 is (-2+4)/2 = 2/2=1. So (1,1) is the center of the circle.</li>
</ol>
<p>If you have trouble seeing this, draw a picture as teteatete has described.</p>
<ol>
<li>This one is algebraically difficult (it is question 20 after all), so I would pick numbers. For example, let’s let x = 10 and y = 5. Now, the call costs 5.55. Subtract off 10 cents for the first minute. This give 5.45. The rest of the minutes are 5 cents, so divide 5.45/.05 = 109. But don’t forget about that first minute you subtracted off. So the answer is 110 minutes. Put a nice big circle around the number 110, and plug x=10 and y=5 into all 5 answer choices. Cross out any answers that do not equal 110. Most likely you will be left with only one answer.</li>
</ol>
<p>Remark: If you are left with more than one correct answer you need to pick new values for x and y and plug in again (but you don’t have to worry about the choices you already eliminated).</p>
<p>Oooh, I just noticed that there is a second question.</p>
<p>In any case, you have to consider that the quantity of minutes is the cost of the call divided by the cost per minute.</p>
<p>For this problem, the cost is 555 and the quantity per minute is y. Therefore, the total number of minutes will be 555/y. But wait. There’s a catch. Before finding the rate you must first deduct the cost of the first minute, x, from the cost.</p>
<p>555-x/y. Now let’s assume x = 55 an y = 10, just to check our work. Indeed, this phone call would last 51 mintutes given those rates. 55+50(10) = 555.</p>
<p>The three points problem has been constructed to provide at shortcuts that allow the student to avoid the more time consuming general solution finding the center of a circle circumscribing a triangle.</p>
<p>The general solution requires solving the circle equation for the three points or finding the intersection of perpendicular bisectors of two sides of the triangle (equivalent methods).</p>
<p>Exactly one circle can be drawn so it passes through three non-colinear points. If your drawing is sufficiently accurate and if the numbers are favorable, estimating the location of the center may suggest a solution (practical for multiple choice problems).</p>
<p>This problem presents a right triangle with a horizontal hypotenuse. (We recognize a right angle at G since the slope of the FG is 1 and the slope of GH is -).</p>
<p>The quick solution uses the fact that FGH is a right triangle: FH forms the diameter of the circle; This is only true because FGH is a right triangle. The center of the circle is at the midpoint of the diameter. If you see this, the problem is easily solved.</p>
<p>The traditional analytic solution requires solving the circle equation: (x-h)^2 + (y-k)^2 = r^2.
This solution works for any set of three points (no right angle required).
We don’t know r or r^2, but we can plug in the three points and equate.
F is at (-2,1)
G is at (1,4)
H is at (4,1)</p>
<p>Plugging in G and H and equating yields a linear equation in h and k…
(1-h)^2 + (4-k)^2 = (4-h)^2 + (1-k)^2
-2h - 8k = -8h -2k
h=k (The equation of a line - in fact the perpendicular bisector of GH)</p>
<p>Plugging in F and H and equating yields a second linear equation in h and k…
(-2-h)^2 + (1-k)^2 = (4-h)^2 + (1-k)^2
4 + 4h = 16 -8h
12h = 12
h=1 (A vertical line - in fact the perpendicular bisector of FH)</p>
<p>The intersection of these lines is (1,1).
You probably wouldn’t want to grind that out during an exam.</p>
<p>(Traditional geometric solution: Derive equations for two perpendicular bisectors and then find the intersection. This method is easy in this case since the perpendicular bisector of GH is line with slope 1 passing through the origin and the perpendicular bisector of FH is a vertical line at x=1. Quick and easy.)</p>
<p>Plug and check is reasonably quick here.</p>
<p>If the triangle has no right angle, we must use the traditional method.
Consider F now at (-4,1), G (1,4), H (4,1).
Where is the circle centered?
(An easy analytic solution exists.)</p>
<p>What about F now at (-4,-1), G (1,4), H (4,1).
Where is the circle centered?
(An easy plug and chug solution exists.)</p>
<ol>
<li><p>If you draw the points, you will see that FGH is a right triangle with hypotenuse FH. Therefore the center of the circle is on the midpoint of FH, which is (1,1), B.</p></li>
<li><p>Let m be the length of the call, in minutes. We have</p></li>
</ol>
<p>555 = x(1) + y(m-1) = x + ym - y</p>
<p>Solving for m, we obtain m = (555 - x + y)/y, answer choice C.</p>