Find the standard form for the equation of a circle with a diameter that has endpoints (0,-3) and (1,2)

Question

(x-h)^2+(y-k)^2=r^2

Answer 1

The diameter has endpts (0, -3) and (1,2)

 

Two things we can get from the above is:

1) the diameter D, and hence the Radius  (by using distance formula)

2) center pts, h, k using the midpt formula

 

Diameter D = SQRT [ (x2 - x1)² + (y2 - y1)² ]

 

D = SQRT [ (1-0)²  + (2 - (-3))² ]

D = SQRT [  1 + 5² ]

D = SQRT(26) 

 

So the radius R = D/2   = SQRT(26)/2

So R² = 26/4 = 13/2

 

CENTER PT is just the MIDPT 

 

Xmid = (x1 + x2)/2

Ymid = (y1 + y2)/2 

 

xmid = (0 + 1)/2   ymid = (-3 + 2)/2 

 

Center pt = (1/2, -1/2) = h,k 

 

Circle eqn (X - h)² + (Y - k)² = R²   is then:  

 

     (X - 1/2)² + (Y + 1/2)² = 13/2 

 

Answer 2

Find the midpoint of the diameter. This will be the center of the circle.

 

The formula for the midpoint is ((x₁ + x₂)/2 , (y₁ + y₂)/2)

 

(0 + 1)/2 = 1/2 or .5

 

(-3 + 2)/2 = -1/2 or -.5

 

Now you need the length of the radius or find the length of the diameter and take half.

 

We'll do the second of those

 

distance or length = √(x₁ - x₂)² + (y₁ - y₂)²

 

   = √( 0 - 1)² + (-3 -2)²

 

   =√1 +25

 

  =√26

 

if that's the diameter then the radius = 1/2 ·√26

 

using the formula where h = .5 and k = -.5

 

(x - .5)² + (y + .5)² = (1/2·√26)²

 

 

(x - .5)² + (y + .5)² = 26/4 = 13/2 = 6.5

 

or with fractions instead of decimals

 

(x - 1/2)² + (y + 1/2)² = 13/2

 

Answer 3

First find the length of the diameter.

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

d = √[(1 - 0)² + (2 + 3)²]

d = √(1 + 25) = √26

r = (1/2)√26     r² = (1/4)(26) = 13/2

Next find the midpoint of the diameter.

x_m = x₁ + x₂      y_m = y₁ + y₂

              2                         2

x_m = 1        y_m = 5     C(1/2, 5/2)

         2                 2

(x - 1/2)² + (y - 5/2)²  = 13/2