Answer:
The equation of the circle is
.
Step-by-step explanation:
The radius of required circle is 2 units. The standard equation of a circle is
.... (1)
where (h,k) is center of the circle and r is radius.
The given equation of a circle is
![x^2+y^2-8x-6y+24=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-8x-6y%2B24%3D0)
![(x^2-8x)+(y^2-6y)+24=0](https://tex.z-dn.net/?f=%28x%5E2-8x%29%2B%28y%5E2-6y%29%2B24%3D0)
If we have the expression
, then we have to add
to make the expression a perfect square.
![(x^2-8x+4^2)+(y^2-6y+3^2)-4^2-3^2+24=0](https://tex.z-dn.net/?f=%28x%5E2-8x%2B4%5E2%29%2B%28y%5E2-6y%2B3%5E2%29-4%5E2-3%5E2%2B24%3D0)
![(x-4)^2+(y-3)^2-1=0](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-3%29%5E2-1%3D0)
.... (2)
From (1) and (2), we get
![h=4,k=3,r=1](https://tex.z-dn.net/?f=h%3D4%2Ck%3D3%2Cr%3D1)
It means the center of this circle is (4,3). So, the center of required circle is also (4,3).
The center is (4,3) and radius is 2, therefore the required equation is
![(x-4)^2+(y-3)^2=2^2](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-3%29%5E2%3D2%5E2)
It can also written as
![(x-4)^2+(y-3)^2=4](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-3%29%5E2%3D4)
Therefore the equation of the circle is
.