How do you write an equation for a circle with (-10 , 0) to (-16 , -10) as a diameter?

1 Answer
Oct 26, 2016

Answer:

The equation is #(x + 13)^2 + (y + 5)^2 = 34#

Explanation:

Start by finding the length of the diameter using the distance formula.

#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#d = sqrt((-6)^2 + (-10)^2)#

#d = sqrt(136)#

#d = 2sqrt(34)#

We will now find the radius using the formula #d = 2r#

#2sqrt(34) = 2r#

#sqrt(34) = r#

Now, let's find the centre of the circle. We can do this using the midpoint formula.

Let #c# be the centre:

#c = ((x_1 + x_2)/2, (y_1 + y_2)/2)#

#c = ((-10 + (-16))/2, (-10 + 0)/2)#

#c = (-26/2, -10/2)#

#c = (-13, -5)#

We can now write the equation of the circle. The form #(x - a)^2 + (x- b)^2 = r^2# represents the equation of the circle, where the point #(a, b)# is the centre and #r# is the radius.

Substituting, we get:

#(x - (-13))^2 + (y - (-5))^2 = (sqrt(34))^2#

#(x + 13)^2 + (y + 5)^2 = 34#

Hopefully this helps!