# Find the standard form for the equation of a circle (x−h)2 +(y−k)2 = r2

+1 vote
Find the standard form for the equation of a circle (x−h)2 +(y−k)2 = r2
with a diameter that has endpoints (0,7) and (1,0).
h=? k=? r=?

Thank You so much!

standard form equation is (x-h)2+(y-k)2=r2, where (x,y) is point on circle, (h,k) is center of circle and r = radius of circle.

so when end points of diameter are given we know that mid point of end points is center of circle.
so (h,k) = [(0+1)/2 , (0 + 7)/2] = ( 1/2, 7/2
)

Now radius is distance of center from one of the point of circle
Consider any one point of the two given end points (0, 7) of diameter and take the distance from center (h,k) = (1/2, 7/2)

we get r by distance formula = [(7/2- 7) ^2 + (1/2 - )^2] = [49/4 + 1/4] = √50/4=25/2

Write the equation in standard form
(x-h)2+(y-k)2=r2 = (x-(1/2))^2 + (y -(7/2))^2 = (√25/2)^2 = 25/2

(x - h)2 + (y - k)2 = r2

In the above equation

We have (x,y) is point on circle

(h,k) is center of circle

We know that mid point of end points is center of circle

So (h,k) = [(0+1)/2 , (7+0)/2] = [ 1/2, 7/2] -- obtained using mid point formula

Now radius is distance between center to one point of the circle

If we take one of the two given end points (1, 0) of diameter and take the distance from center (h,k) = (1/2, 7/2)

We get r by distance formula = √ [(1 - 1/2)2 + (0 - 7/2)2]

= √ [(1/2)2 + (-7/2)2]

= √ [1/4 + 49/4]

= √(50/4)

Hence we have the equation in standard form

(x-h)2+(y-k)2 = r2

(x-(1/2))2 + (y -(7/2))2 = [(√50)/2]2

(x-(1/2))2 + (y -(7/2))2 = 50/4

The solutions is (x-(1/2))2 + (y -(7/2))2 = [25/2]