<p>In the coordinate plane, the points F(-2, 1), G(1,4), and H(4,1) lie on a circle with center P. What are the coordinates of P?</p>
<p>A (0,0)
B (1,1) correct ans
C (1,2)
D(1,-2)
E (2.5, 2.5)</p>
<p>I was thinking of using the distance formula to check all the points but its really time consuming... there must be a faster way to do this right??</p>
<p>You just need to find the midpt of F and H. The midpoint is 1,1. You’ll find that the x-coord is 3 away from both, so the radius is 3, thus, your y-coord is also 1 (3 away from 1)</p>
<p>I don’t know if the 2nd part of that is necessary (since I had the answer after one midpt taking, but do it just incase)</p>
<p>It’s best to visually see these types of problems: Sketch it out!</p>
<p>I was thinking of that too, it is possible to find that the mid point of F and H is (1,1) thus the center’s x coordinate must be 1. but how can you be so sure that the radius is the distance from point F and point H to their mid pt (1,1)? i thought that F and H would be similar distances from any pt along the line x=1?
… hope that wasn’t too confusing.</p>
<p>I see your point – they don’t specify that the two points are endpoints of a diameter. But if you draw it out, you see that the third point they gave you is also 3 units away from the midpoint of the other two. So that midpoint is in fact the center of the circle. (If you know the compass construction for finding the center of the circle that circumscribes a triangle, it works by the same principle: you find the perpendicular bisectors of each of the three sides. Where they meet is the circumcenter.)</p>
<p>A less mathematical, more cynical answer is that I have never seen an SAT question that required the use of the distance formula. It seems that when you draw the diagram, there is always some “lucky” coincidence in the design of the problem that lets you get away with just counting the boxes.</p>