# Find the standard form of the given equation of the circle that has a diameter wih the given endpoints (5,-1) , (5,7)

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Let the equation be:

`(x-a)^2 + (y-b)^2 = r^2``` such that: (a,b) is the center and r is the radius.

First we will find the radius.

The length of the diameter is given by :

D = `sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`

`==> D= sqrt((5-5)^2 +(7+1)^2)= sqrt(8^2)= 8`

`==> r= 8/2 = 4`

`==> (x-a)^2 + (y-b)^2 = 16`

`` Now we will substitute with the points (5,-1) and (5,7) because they are on the graph of the circle, then they must verify the equation of the circle.

==> `(5-a)^2 + (-1-b)^2 = 16`

`==> (5-a)^2 + (7-b)^2 = 16`

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```` We will subtract the equations:

`==gt (-1-b)^2 - (7-b)^2 = 0 `

`==gt (-1-b)^2 = (7-b)^2 `

`==gt 1+2b +b^2 = 49 -14b +b^2 `

`==gt 16b = 48 `

`==gt b= 48/16 = 3`

`==> b= 3`

`==> (5-a)^2 + (7-3)^2 = 16`

`==> (5-a)^2 + 16 = 16`

`==> (5-a)^2 = 0`

`==> a = 5`

`==> (a, b)= (5,3) , r= 4`

Then, the equation of the circle is :

`=> (x-5)^2 + (y-3)^2 = 16`

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