Find the equation of the circle, in standard form, where (1, Solution We have to find the circle which has ( 1, -5) and ( -3 , 6) as the end points of a diameter. Now, the distance between the two points is sqrt [( 1 +3)^2 + (-5-6)^2] => sqrt [ 4^2 + 11^2] => sqrt ( 16 + 121) => sqrt 137. The radius is (sqrt 137)/2. The mid point between the two point given is ( -2/2 , 1/2) The equation of the circle is ( x +1)^2 + (y - 1/2)^2 = (sqrt 137)^/4 => x^2 + 2x + 1 + y^2 + 1/4 - y = 137 / 4 => x^2 + y^2 + 2x - y + 5/4 = 137 / 4 => x^2 + y^2 + 2x -y = 132 / 4 => => x^2 + y^2 + 2x -y = 33 => x^2 + y^2 + 2x -y -33 = 0 The equation of the circle is x^2 + y^2 + 2x + -y -33 = 0.