Question

How do you write an equation for a circle with endpoints of a diameter are at (1,1) and (1, -9)?

Answer 1

`(x-1)^2+(y+4)^2=5^2`. Expansion gives `x^2+y^2-2x+8y-8=0`

Explanation:

The center bisects the diameter. So it is the midpoint

`C((1+1)/2, (1-9)/2))=(1, -4).`

The radius a is half the diameter #=(1/2)sqrt(0^2+10^2)=(1/2)10=5.

So, the equation of the circle is `(x-1)^2+(y+4)^2=5^2`

Expansion gives `x^2+y^2-2x+8y-8=0`