The center of the circle is (h,k) = (-2,-3)
The radius of the circle is r = 2
The standard form of equation of the circle is
[tex]{(x + 2)}^{2} + {(y + 3)}^{2} = 4[/tex]
How to find the center, radius and standrad form of the circle?
The general form of equation of the circle is
[tex] {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} [/tex]
Here, (h,k) means centre of the circle.
r means radius of the circle.
given that coordinate points of centre of circle is (-2,-3).
Hence the (h,k) = (-2,-3)
How to find the radius of the circle?
Now to find the radius of the circle
The distance from a circle's centre to its circumference is its radius.
The distance from a circle's centre (-2,-3) to its circumference (0,-3) is its radius.
using the formula, distance between the two points to obtain radius.
[tex]d = \sqrt{(x1 - x2) {}^{2} + {(y1 - y2)}^{2} } \\ r = \sqrt{ {( - 2 - 0)}^{2} + {( - 3 - ( - 3))}^{2} } \\ r = \sqrt{ {( - 2)}^{2} + {( - 3 + 3)}^{2} } \\ r = \sqrt{ {4}^{2} + 0 } \\ r = \sqrt{4} \\ r = 2[/tex]
How to find the standard form of equation of the circle?
(h,k) = (-2,-3)
r = 2
subtitue the (h,k) and r values to get the standard form of equation of the circle.
[tex](x - h) {}^{2} + {(y - k)}^{2} = {r}^{2} [/tex]
[tex] {(x - ( - 2))}^{2} + {(y - ( - 3))}^{2} = {r}^{2} [/tex]
[tex] {(x + 2)}^{2} + {(y + 3)}^{2} = 4[/tex]
Learn more about circle, refer:
https://brainly.com/question/24810873
#SPJ9
