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solmaris [256]
2 years ago
10

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

Mathematics
2 answers:
RUDIKE [14]2 years ago
7 0
(x-2)^2+(y+3)^2=32
do you need to show work?
Svet_ta [14]2 years ago
4 0

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

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