Question

Find an equation of the circle whose diameter has endpoints (-6, -1) and (-2,3).

Answer 1

Explanation:

Let find the distance of the diameter. using distance formula.

`(3 + 1) {}^(2) + ( - 2 + 6) {}^(2) = √(8)`

The diameter is sqr root of 8 units.

A circle equation is

`{x}^(2) + {y}^(2) = {r}^(2)`

where r is the radius. The radius is half the diameter so

`r = ( √(8) )/(2) = ( √(8) )/( √(4) ) = √(2)`

`{r}^(2) = { √(2) }^(2) = 2`

So our radius is 2.

Now we need to find the midpoint or Center of the diameter.

`( - 6 - 2)/(2) = - 4`

`(3 - 1)/(2) = 1`

So the center of the circle is (-4,1). So our equation of the Circle us

`(x + 4) {}^(2) + (y - 1) {}^(2) = ( √(2) ) {}^(2)`