# The endpoints of the diameter of a circle are (-4,-5) and (-2,-1). What is the center, radius, and equation?

##### 1 Answer
Aug 27, 2016

The Centre is$\left(- 3 , - 3\right) , \text{radius r} = \sqrt{5}$.

The eqn.$: {x}^{2} + {y}^{2} + 6 x + 6 y + 13 = 0$

#### Explanation:

Let the given pts. be $A \left(- 4 , - 5\right) \mathmr{and} B \left(- 2 , - 1\right)$

Since these are the extremities of a diameter, the mid-pt. $C$ of segment $A B$ is the centre of the circle.

Hence, the centre is $C = C \left(\frac{- 4 - 2}{2} , \frac{- 5 - 1}{2}\right) = C \left(- 3 , - 3\right)$.

$r \text{is the radius of the circle} \Rightarrow {r}^{2} = C {B}^{2} = {\left(- 3 + 2\right)}^{2} + {\left(- 3 + 1\right)}^{2} = 5$.

$\therefore r = \sqrt{5}$.

Finally, the eqn. of the circle, with centre $C \left(- 3 , - 3\right)$, and radius$r$, is

${\left(x + 3\right)}^{2} + {\left(y + 3\right)}^{2} = {\left(\sqrt{5}\right)}^{2} , i . e . , {x}^{2} + {y}^{2} + 6 x + 6 y + 13 = 0$