Question 261538
The equation of a circle is:
{{{(x - h)^2 + (y-k)^2 = r^2}}}
where (h, k) is the center of the circle and r is the radius. So we need the center and the radius of the circle to write the equation.<br>
The center of a circle is always the midpoint of a diameter. So let's use the midpoint formula to find the center of the circle.
{{{x[m] = (x[1] + x[2])/2}}} and {{{y[m] = (y[1] + y[2])/2}}}
{{{x[m] = (-4 + 6)/2}}} and {{{y[m] = (3 + (-8))/2}}}
{{{x[m] = 2/2}}} and {{{y[m] = (-5)/2}}}
{{{x[m] = 1}}} and {{{y[m] = (-5)/2}}}
So the center of the circle is (1, -5/2)<br>
The radius of a circle is one-half the length of the diameter. We have the coordinates of the diameter so we can use the distance formula to find its length. And half of this length will be the radius we need. The distance formula is:
{{{d = sqrt((x[2] - x[1])^2 + (y[2] - y[1])^2)}}}
Substituting the coordinates into this we get:
{{{d = sqrt((6 - (-4))^2 + (-8 - 3)^2)}}}
which simplifies as follows:
{{{d = sqrt((10)^2 + (-11)^2)}}}
{{{d = sqrt(100 + 121)}}}
{{{d = sqrt(221)}}}
This is the length of the diameter. The radius will be 1/2 of this:
{{{r = sqrt(221)/2}}}<br>
Now that we have the center of the circle and the length of the radius, we can substitute these values into {{{(x - h)^2 + (y-k)^2 = r^2}}}:
{{{(x - 1)^2 + (y - (-5/2))^2 = (sqrt(221)/2)^2}}}
which simplifies to:
{{{(x - 1)^2 + (y + 5/2)^2 = 221/4}}}