equation of a cirlce

Question

find the equation of a circle (standard form) with diameters whose endpoints are (2,5) and (7,-1)?

Answer 1

Let the points are (x₁,y₁) = (2, 5) and (x₂, y₂) = (7, - 1).

Standard form for circle : (x - h )^2 + (y - k )^2 = r ^2.

The midpoint of the diameter is centre(h,k ).

mid point (h,k ) = [(x₁ + x₂)/2, (y₁ + y₂)/2].

                       = [(2 + 7)/2, (5 - 1)/2]

                       = [9/2, 4/2]

                       = [9/2, 2].

Centre(h,k )  = (9/2, 2).

The distance from the centre to a another point on the circle is the radius.

Radius r = sqrt [ (h - x₁)^2 + (k - y₁)^2 ]

            = √ [ (9/2 - 2)^2 + (2 - 5)^2 ]

            = √ [ (5/2)^2 + (- 3)^2 ]

            = √ [ 25/4 + 9 ]

            = √61/4.

Radius r = √61/4 units.

Substitute the values of (h,k )  = (9/2, 2) and r = √61/4 in standard form for circle.

(x - 9/2)^2 + (y - 2)^2 = (√61/4)^2

(x - 9/2)^2 + (y - 2)^2 = 61/4.

The equation of a circle in standard form is (x - 9/2)^2 + (y - 2)^2 = 61/4.