Given:
Center of a circle is (3,9).
Solution point is (-2,21).
To find:
The standard form of the circle.
Solution:
Radius is the distance between center (3,9) and the solution point (-2,21).
![r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![r=\sqrt{(-2-3)^2+(21-9)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%28-2-3%29%5E2%2B%2821-9%29%5E2%7D)
![r=\sqrt{(-5)^2+(12)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%28-5%29%5E2%2B%2812%29%5E2%7D)
![r=\sqrt{25+144}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B25%2B144%7D)
On further simplification, we get
![r=\sqrt{169}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B169%7D)
![r=13](https://tex.z-dn.net/?f=r%3D13)
The radius of the circle is 13 units.
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 and r is the radius.
Putting h=3, k=9 and r=13, we get
![(x-3)^2+(y-9)^2=(13)^2](https://tex.z-dn.net/?f=%28x-3%29%5E2%2B%28y-9%29%5E2%3D%2813%29%5E2)
![(x-3)^2+(y-9)^2=169](https://tex.z-dn.net/?f=%28x-3%29%5E2%2B%28y-9%29%5E2%3D169)
Therefore, the standard form of the circle is
.