Find the equation of the circle when the endpoints of diameter are \[A\left( { - 2, - 3} \right)\] and \[B\left( { - 3, - 5} \right).\]

Hint: Centre is the midpoint of diameter and distance from centre to either of the endpoints of diameter is radius. Thereafter, we will use the midpoint formula in the given values. Further, we will use distance formula to find the equation of the circle given below:
\[x = \dfrac{{{x_1} + {x_2}}}{2},\,y = \dfrac{{{y_1} + {y_2}}}{2}\] (midpoint formula)
\[{(x - {x_1})^2} + {(y - {y_1})^2} = {r^2}\] (Centre radius form of circle)
\[ \Rightarrow r = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \] (Distance formula)

Complete step by step answer:
(1) Given diametric ends of the circle are \[A\left( { - 2, - 3} \right)\] and \[B\left( { - 3, - 5} \right)\]

(2) Using the midpoint formula, we calculate the coordinate of the centre first.
\[x = \dfrac{{{x_1} + {x_2}}}{2},y = \dfrac{{{y_1} + {y_2}}}{2}\]
Where \[{x_1} = - 2,\,\,{y_1} = - 3,\,\,{x_2} = - 3\,\,\,and\,\,{y_2} = - 5\]

(3) Using these values in formula $(2)$ we have
\[x = \dfrac{{( - 2) + ( - 3)}}{2}\]
\[x = \dfrac{{ - 5}}{2}\]
\[y = \dfrac{{( - 3) + ( - 5)}}{2}\]
\[y = \dfrac{{ - 8}}{2}\]
\[y = - 4\]
\[\therefore \] Coordinate of centre is\[\left( {\dfrac{{ - 5}}{2}, - 4} \right)\]

(4) We apply distance formula between centre of the circle and either end of diameter to calculate its radius:
\[D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Here, \[{x_2} = - 3,\,\,\,{y_2} = - 5\]
\[{x_1} = - \dfrac{5}{2},\,\,{y_1} = - 4\]
Using in above formula we have
\[D = \sqrt {{{\left( { - 3 - {{\left( {\dfrac{{ - 5}}{2}} \right)}^{}}} \right)}^2} + {{( - 5 - ( - 4))}^2}} \]
\[ = \sqrt {{{\left( { - 3 + \dfrac{5}{2}} \right)}^2} + {{( - 5 + 4)}^2}} \]
\[ = \sqrt {\left( {\dfrac{{ - 6 + 5}}{2}} \right) + {{( - 1)}^2}} \]
\[ = \sqrt {{{\left( {\dfrac{{ - 1}}{2}} \right)}^2} + {{( - 1)}^2}} \]
\[ = \sqrt {\dfrac{1}{4} + 1} \]
\[ = \dfrac{{\sqrt {1 + 4} }}{4}\]
\[ = \sqrt {\dfrac{5}{4}} \]

(5) Therefore, radius of the circle is \[\sqrt {\dfrac{5}{4}} \]
\[ \Rightarrow |OA|\, = \,|OB|\, = \sqrt {\dfrac{5}{4}} \]

(6) Now, we use the centre radius formula to find the equation of the circle.
\[{(x - {x_1})^2} + {(y - {y_1})^2} = {r^2}\]
Where (x1, y1) is either of the diametric end and $r$ be the radius of the circle.

(7) Using value of \[({x_1},{y_1}) = ( - 3, - 5)\,\,and\,\,\,r = \sqrt {\dfrac{5}{4}} \] in above formula.
\[{\left[ {x - ( - 3)} \right]^2} + {\left[ {y - ( - 5)} \right]^2} = {\left( {\sqrt {\dfrac{5}{4}} } \right)^2}\]
\[ \Rightarrow {(x + 3)^2} + {(y + 5)^2} = \left( {\dfrac{5}{4}} \right)\]
Using algebraic identity: \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[ \Rightarrow {x^2} + 9 + 6x + {y^2} + 25 + 10y = \dfrac{5}{4}\]
\[ \Rightarrow 4\left( {{x^2} + 9 + 6x + {y^2} + 25 + 10y} \right) = 5\]
\[ \Rightarrow 4{x^2} + 4{y^2} + 24x + 40y + 136 = 5\]
\[ \Rightarrow 4{x^2} + 4{y^2} + 24x + 40y + 136 - 5 = 0\]
\[ \Rightarrow 4{x^2} + 4{y^2} + 24x + 40y + 131 = 0\]

Additional Information: A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the centre, equivalently it is the curve traced out by a point that moves in a plane so that its distance from a given point is constant.

Note: We can also use formula \[\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0\] to find equation of circle.
Here, \[\left( {{x_1},{y_1}} \right){\text{ }}and{\text{ }}\left( {{x_2},{y_2}} \right)\] are diametric ends.