Answer:
[tex](x-4)^2+(y-3)^2=2^2[/tex]
Step-by-step explanation:
The given circle has equation; [tex]x^2+y^2-8x-6y+24=0[/tex]
Comparing to the general equation of the circle: [tex]x^2+y^2+2ax+2by+c=0[/tex]
We have [tex]2a=-8\implies a=-4[/tex] and [tex]2b=-6\implies b=-3[/tex]
The center of this circle is (-a,-b)=(4,3).
The required circle has radius r=2 units.
The equation of a circle, given the center (h,k) and radius r, is given by:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
We substitute the values to obtain [tex](x-4)^2+(y-3)^2=2^2[/tex]
