Question

match each equation to its description.
a: circle centered at (1,0) with a radius of 5
b: circle centered at (1,−6) with a radius of 5
c: circle centered at (1,−6) with a radius of √5
d: circle centered at (0,−6) with a radius of 5
e: circle centered at (1,6) with a radius of √5
1: (x−1)²+(y+6)²=25 2: (x−1)²+(y+6)²=5 3: (x−1)²+y²=25
4: (x−1)²+(y−6)²=5 5: x²+(y+6)²=25

Explanation:

Step1: Recall circle standard form

The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius.

Step2: Match A to equation

Center $(1,0)$, $r=5$: $(x-1)^2 + (y-0)^2 = 5^2$ → $(x-1)^2 + y^2 = 25$ (Equation 3)

Step3: Match B to equation

Center $(1,-6)$, $r=5$: $(x-1)^2 + (y-(-6))^2 = 5^2$ → $(x-1)^2 + (y+6)^2 = 25$ (Equation 1)

Step4: Match C to equation

Center $(1,-6)$, $r=\sqrt{5}$: $(x-1)^2 + (y+6)^2 = (\sqrt{5})^2$ → $(x-1)^2 + (y+6)^2 = 5$ (Equation 2)

Step5: Match D to equation

Center $(0,-6)$, $r=5$: $(x-0)^2 + (y+6)^2 = 5^2$ → $x^2 + (y+6)^2 = 25$ (Equation 5)

Step6: Match E to equation

Center $(1,6)$, $r=\sqrt{5}$: $(x-1)^2 + (y-6)^2 = (\sqrt{5})^2$ → $(x-1)^2 + (y-6)^2 = 5$ (Equation 4)

Answer:

A: circle centered at (1, 0) with a radius of 5 → 3: $(x-1)^2 + y^2 = 25$
B: circle centered at (1, -6) with a radius of 5 → 1: $(x-1)^2 + (y+6)^2 = 25$
C: circle centered at (1, -6) with a radius of $\sqrt{5}$ → 2: $(x-1)^2 + (y+6)^2 = 5$
D: circle centered at (0, -6) with a radius of 5 → 5: $x^2 + (y+6)^2 = 25$
E: circle centered at (1, 6) with a radius of $\sqrt{5}$ → 4: $(x-1)^2 + (y-6)^2 = 5$