Question

Determine the equation of a circle whose diameter has the endpoints (-1, 2) and (7.

Answer 1

Answer:

option 2

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

The centre is at the midpoint of the endpoints of the diameter.

Using the midpoint formula with (- 1, 2) and (7, - 4), then

centre = ( $\frac{-1+7}{2}$ , $\frac{2-4}{2}$ ) = (3, - 1 )

The radius is the distance from the centre to either of the endpoints of the diameter.

Using the distance formula

r = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (3, - 1) and (x₂, y₂ ) = (- 1, 2)

r = $\sqrt{(3+1)^2+(2+1)^2}$

  = $\sqrt{4^2+3^2}$

  = $\sqrt{16+9}$ = $\sqrt{25}$ = 5

Thus

(x - 3)² + (y - (- 1))² = 5² , that is

(x - 3)² + (y + 1)² = 25 ← equation of circle

Answer 2

Answer:

(x - 3)^2 + (y + 1) = 25

Step-by-step explanation:

First find the midpoint of the diameter, because that represents the center of the circle.

                                                                     -1 + 7

The x-coordinate of the midpoint is xm = ---------- = 3

                                                                          2

                                                 2 - 4

and the y-coordinate is ym = ---------- = -1

                                                     2

And so the center of this circle is at (3, -1).

Use the Pythagorean Theorem to determine the square of the radius:

square of radius = 4^2 + (-3)^2 = 16 + 9 = 25

And so the equation of this circle is (x - 3)^2 + (y + 1) = 25