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).
![Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=Distance%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![r=\sqrt{(-5-(-2))^2+(3-1)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%28-5-%28-2%29%29%5E2%2B%283-1%29%5E2%7D)
![r=\sqrt{(-5+2)^2+(2)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%28-5%2B2%29%5E2%2B%282%29%5E2%7D)
![r=\sqrt{(-3)^2+(2)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%28-3%29%5E2%2B%282%29%5E2%7D)
On further simplification, we get
![r=\sqrt{9+4}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B9%2B4%7D)
![r=\sqrt{13}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B13%7D)
The standard form of a circle is:
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
Where, (h,k) is the center of the circle and r is the radius of the circle.
Substitute h=-2, k=1 and
.
![(x-(-2))^2+(y-1)^2=(\sqrt{13})^2](https://tex.z-dn.net/?f=%28x-%28-2%29%29%5E2%2B%28y-1%29%5E2%3D%28%5Csqrt%7B13%7D%29%5E2)
![(x+2)^2+(y-1)^2=13](https://tex.z-dn.net/?f=%28x%2B2%29%5E2%2B%28y-1%29%5E2%3D13)
Therefore, the equation of the circle is
.