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 x² + y² - 8x - 6y + 24 = 0. (x + 4)² + (y + 3)² = 2 (x - 4)² + (y - 3)² = 2 (x - 4)² + (y - 3)² = 2² (x + 4)² + (y + 3)² = 2²

Explanation:

Step1: Rewrite the given circle equation in standard form

The general equation of a circle is $x^{2}+y^{2}+Dx + Ey+F = 0$, and its standard - form is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where the center is $(a,b)$ and the radius is $r$. Given $x^{2}+y^{2}-8x - 6y + 24 = 0$. Complete the square for $x$ and $y$ terms. For the $x$ - terms: $x^{2}-8x=(x - 4)^{2}-16$. For the $y$ - terms: $y^{2}-6y=(y - 3)^{2}-9$. Then $x^{2}+y^{2}-8x - 6y + 24=(x - 4)^{2}-16+(y - 3)^{2}-9 + 24=0$. Rearranging gives $(x - 4)^{2}+(y - 3)^{2}=1$. The center of the circle $x^{2}+y^{2}-8x - 6y + 24 = 0$ is $(4,3)$.

Step2: Determine the equation of the new circle

We know the radius of the new circle $r = 2$ and the center $(a,b)=(4,3)$. The standard - form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$. Substituting $a = 4$, $b = 3$, and $r = 2$ into the equation, we get $(x - 4)^{2}+(y - 3)^{2}=2^{2}$.

Answer:

$(x - 4)^{2}+(y - 3)^{2}=2^{2}$