Question 623929
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Hi there,

The problem:
Find the equation of the circle whose diameter has endpoints (-9, -2) and (1, 1). Write it in the form
(x-h)^2 + (y-k)^2 = r^2. Identify the center of the circle (h.k) and its radius r.

A Solution:
This problem has a lots parts. I can see why you might be feeling frustrated. We'll take this step 
by step.

We have two points on the circle that are endpoints of one of its diameters. A diameter always 
passes through the center of the circle. In fact, the center is the midpoint of the diameter.

We can use the midpoint formula to find the center of the circle. The midpoint formula is
M(x,y) = ((x[1]+x[2])/2 , (y[1]+y[2])/2)

The center C of this circle is located at the point (h,k) midway between (x[1],y[1]=(1,1) and 
(x[2],y[2])=(-9,-2). Using the midpoint formula, we have

C(h,k) = ((-9+1)/2 , (-2+1)/2)

Simplifying, we get,
C(h,k) = (-4,-1/2)

The center of the circle is at the point (-4,-1/2). Using the vertex form for the equation for a circle,
substitute these values for h and k.
{{{(x-h)^2 + (y-k)^2 = r^2}}}
{{{(x-(-4))^2 + (y-(-1/2))^2 = r^2}}}

Simplify.
{{{(x+4)^2 + (y+1/2)^2 = r^2}}}

We know that those endpoints to the diameter are actually points on the circle, so we can substitute
one of them into the equation for x and y. Then we'll be able to solve for r^2. Either point will work;
let's use (1,1).
{{{((1)+4)^2 + ((1)+1/2)^2 = r^2}}}

Simplify and solve for r^2.
{{{(5)^2 + (3/2)^2 = r^2}}}
{{{25+9/4=r^2}}}
{{{109/4=r^2}}}

We have (h,k) and r^2, so we can write our whole equation.
{{{(x+4)^2+(y+1/2)^2=109/4}}}

This is the vertex form of the equation, but the radius is r not r^2. So we take the square root of 
109/4 to get the radius.
{{{sqrt(109/4)=(sqrt(109))/2=(1/2)*(sqrt(109))}}}

So, the center of your circle is (h,k)=(-4, -1/2) and the radius of the circle is r = (1/2)*sqrt(109)

Please email me if you have questions about any of this. I'd appreciate the feedback and it will help 
me become a better "explainer."

Ms.Figgy
mth.in.the.vortex@gmail.com

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