Question

Find the equation of a circle given only its diameter endpoints: (-2,1) and (6,-7)

Answer 1

(x-2)^2+(y+3)^2=32
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Answer 2

Answer:

(x - 2)² + (y + 3)² = 32

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 of the circle is at the midpoint of the diameter

Calculate the centre (x, y ) using the midpoint formula

(x, y ) = ( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2}$ )

with (x₁, y₁ ) = (- 2, 1) and (x₂, y₂ ) = (6, - 7)

(x , y ) = ( $\frac{-2+6}{2}$ , $\frac{1-7}{2}$ ) = ( $\frac{4}{2}$ , $\frac{-6}{2}$ ) = (2, - 3)

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

Calculate the radius using the distance formula

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

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

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

  = $\sqrt{(-4)^2+(1+3)^2$

   = $\sqrt{16+4^2}$

   = $\sqrt{16+16}$

   = $\sqrt{32}$

Then equation of circle is

(x - 2)² + (y - (- 3) )² = ($\sqrt{32}$ )² , that is

(x - 2)² + (y + 3)² = 32