Given:
The center of the circle = (-2,1).
Circle passes through the point (-5,3).
To find:
The equation of the circle.
Solution:
Radius is the distance between the center of the circle and any point on the circle. So, radius of the circle is the distance between the points (-2,1) and (-5,3).
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]r=\sqrt{(-5-(-2))^2+(3-1)^2}[/tex]
[tex]r=\sqrt{(-5+2)^2+(2)^2}[/tex]
[tex]r=\sqrt{(-3)^2+(2)^2}[/tex]
On further simplification, we get
[tex]r=\sqrt{9+4}[/tex]
[tex]r=\sqrt{13}[/tex]
The standard form of a circle is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where, (h,k) is the center of the circle and r is the radius of the circle.
Substitute h=-2, k=1 and [tex]r=\sqrt{13}[/tex].
[tex](x-(-2))^2+(y-1)^2=(\sqrt{13})^2[/tex]
[tex](x+2)^2+(y-1)^2=13[/tex]
Therefore, the equation of the circle is [tex](x+2)^2+(y-1)^2=13[/tex].