Circles and Distance   The Distance Formula Recall that the Pythagorean Theorem states that if a, b, c are sides of a right triangle with c the hypotenuse, then         a2 + b2 = c2 Let (x,y) be a point in the plane.  Then if we draw the triangle with one vertex at the origin, one at (x,y) and one at (x,0) then we have a right triangle which has one leg of length x and the other of length y.  The length of the hypoteneuse is the distance from the origin to the point (x,y).  By the Pythagorean Theorem, this distance is                    Exercise:  Find the distance from the origin to the point (5,12)         We can generalize this notion to find the distance between two points (x1,y1) and (x2,y2)  We draw the triangle and see that the length of its legs equal         x2 - x1   and    y2 - y1         Therefore the distance from (x1,y1) to (x2,y2) is          Example Find the distance from (2,-3) to (-1,4)   Solution:   We use the distance formula:             Recall that a circle with radius r and center (h,k) is defined by the set of point of distance r from the point (h,k).  If (x,y) is on the circle then the distance from (h,k) to (x,y) is r.  The distance formula tells us that         (x - h)2  + (y - k)2 = r2 This is called the standard form of the equation of the circle.   Example The equation of the circle with radius 3 centered at (1,2) is        (x - 1)2 + (y - 2)2  =  9   Example:  Find the center and radius of the circle         x2 + y2 - 4x + 6y - 12  =  0 We must get the equation into standard form by completing the squares.  We have         x2 - 4x   + y2 + 6y  =  12         -4/2  =  -2      and      6/2  =  3         (-2)2  =  4      and      32  =  9 Adding and subtracting we have         x2 - 4x + 4 - 4 + y2 + 6y + 9 - 9  =  12         (x - 2)2 + (y + 3)2  =  12 + 4 + 9  =  25  =  52 So that the center of the circle is (2,-3) and the radius is 5. Back to the Intermediate Algebra (Math 154) Home Page e-mail Questions and Suggestions