Question

If a circle has diameter endpoints at (-1,7) and (6,2), what is its center and radius

Answer 1

Answer:

The radius is $\frac{\sqrt{74}}{2}$.

The center is (5/2 , 9/2).

Step-by-step explanation:

The radius is half the diameter.  We aren't given the length of the diameter but we are given endpoints to one of them.

So let's find the length of that diameter using the distance formula.

The distance formula is

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.

I just do big x minus small x or big y minus small y.

Anyways the points are (-1,7) and (6,2).

The x distance is 6-(-1)=7.

The y distance is 7-2=5.

So we have this using the distance formula so far:

$d=\sqrt{(6-(-1))^2{(7-2)^2}$

$d=\sqrt{7^2+5^2}$

$d=\sqrt{49+25}$

$d=\sqrt{74}$

So the radius is half that much because that was the distance between the endpoints of a diameter.

So the radius is $\frac{\sqrt{74}}{2}$.

Now the center of a circle will lie on the midpoint of a diameter, any given diameter.

We have the endpoints of one, so we just need to use midpoint formula.

Midpoint formula says the midpoint is (average of x, average y).

Midpoint formula: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

So average of x's is (-1+6)/2=5/2.

The average of y's is (7+2)/2=9/2.

So the midpoint of the diameter or the center of the circle is at (5/2 , 9/2).