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 a^{2} + b^{2} = c^{2} 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 (x_{1},y_{1}) and (x_{2},y_{2}) We draw the triangle and see that the length of its legs equal x_{2}  x_{1} and y_{2}  y_{1}
Therefore the distance from (x_{1},y_{1}) to (x_{2},y_{2}) 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} = r^{2} 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 x^{2} + y^{2}  4x + 6y  12 = 0 We must get the equation into standard form by completing the squares. We have x^{2}  4x + y^{2} + 6y = 12 4/2 = 2 and 6/2 = 3 (2)^{2} = 4 and 3^{2} = 9 Adding and subtracting we have x^{2}  4x + 4  4 + y^{2} + 6y + 9  9 = 12 (x  2)^{2} + (y + 3)^{2} = 12 + 4 + 9 = 25 = 5^{2} So that the center of the circle is (2,3) and the radius is 5. Click here to interactively practice graphing a circle. Back to the Intermediate Algebra (Math 154) Home Page 
