Question 1183676

Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
<pre>If you want to confuse someone or yourself to death, then use the many variables that the other person presented to you. 
If not, then just use the MIDPOINT formula: {{{matrix(3,5, "(", (x[1] + x[2])/2, ",", (y[1] + y[2])/2, ")", "(", (- 2 + 6)/2, ",", (- 1 + - 3)/2, ")", 
"(", 4/2, ",", - 4/2, ")")}}}, to get:
                                            {{{matrix(1,5, "(2,", "- 2),", or, "(h,", "k)")}}} <===== CENTER coordinates of circle
Standard form of the equation of a CIRCLE: {{{matrix(1,3, (x - h)^2 + (y - k)^2, "=", r^2)}}}
                                           {{{matrix(1,3, (- 2 - 2)^2 + (- 1 - - 2)^2, "=", r^2)}}} ------ In the ABOVE, substituting (2, - 2) for (h, k) and (- 2, - 1) for (x, y). NOTE that (6, - 3) could also be used for (x, y)
                                           {{{matrix(3,3, (- 4)^2 + (1)^2, "=", r^2, 16 + 1, "=", r^2, 17, "=", r^2)}}}
For the equation in Standard form, we finally get: {{{highlight_green(matrix(1,3, (x - 2)^2 + (y + 2)^2, "=", 17))}}}