Question

which equation represents the circle described?
the radius is 2 units.
the center is the same as the center of a circle whose equation is ( x^{2}+y^{2}-8x - 6y+24 = 0 ).
( (x + 4)^{2}+(y + 3)^{2}=2 )
( (x - 4)^{2}+(y - 3)^{2}=2 )
( (x - 4)^{2}+(y - 3)^{2}=2^{2} )
( (x + 4)^{2}+(y + 3)^{2}=2^{2} )

Explanation:

Step1: Find the center of the circle \(x^{2}+y^{2}-8x - 6y+24 = 0\)

The general equation of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\), and the general form \(x^{2}+y^{2}+Dx+Ey+F = 0\) has center \((-\frac{D}{2},-\frac{E}{2})\).
For \(x^{2}+y^{2}-8x - 6y+24 = 0\), \(D=-8\), \(E = - 6\).
The center \((a,b)=(4,3)\) (since \(a=-\frac{-8}{2}=4\) and \(b =-\frac{-6}{2}=3\)).

Step2: Use the standard form of the circle equation

The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Given \(r = 2\) and \((h,k)=(4,3)\), the equation is \((x - 4)^{2}+(y - 3)^{2}=2^{2}\).

Answer:

\((x - 4)^{2}+(y - 3)^{2}=2^{2}\) (the third option)