Question

learning goal i can derive the equation of a circle given its center and radius. i can determine the center and radius of a circle given its equation. lesson 4.1 checkpoint once you have completed the above problems and checked your solutions, complete the lesson checkpoint below. complete the lesson reflection above by circling your current understanding of the learning goal. use the graph given to answer questions #1 - 3. 1. what are the coordinates of the circle’s center? (__, ) 2. what is the radius of the circle? r =__ 3. given the circle, write the equation of the circle in standard form. 4. which of the following points is on the circle centered at (8, 9) with a radius of 10 units? a (0, 3) b (-3, 5) c (9, 1) d (-4, 1)

Explanation:

Step1: Locate center on graph

By observing the graph, the circle is centered at the origin. So the coordinates of the center are $(0,0)$.

Step2: Determine radius

Count the grid - units from the center $(0,0)$ to the edge of the circle. The radius $r = 4$.

Step3: Write equation of circle

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Substituting $h = 0,k = 0,r = 4$ gives $(x - 0)^2+(y - 0)^2=4^2$, or $x^{2}+y^{2}=16$.

Step4: Check which point is on the circle

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For a circle centered at $(h,k)=(8,9)$ with radius $r = 10$, a point $(x,y)$ is on the circle if $d=\sqrt{(x - 8)^2+(y - 9)^2}=10$.

• For point A$(0,3)$: $d=\sqrt{(0 - 8)^2+(3 - 9)^2}=\sqrt{64 + 36}=\sqrt{100}=10$.
• For point B$(-3,5)$: $d=\sqrt{(-3 - 8)^2+(5 - 9)^2}=\sqrt{121 + 16}=\sqrt{137}

eq10$.

• For point C$(9,1)$: $d=\sqrt{(9 - 8)^2+(1 - 9)^2}=\sqrt{1+64}=\sqrt{65}

eq10$.

• For point D$(-4,1)$: $d=\sqrt{(-4 - 8)^2+(1 - 9)^2}=\sqrt{144 + 64}=\sqrt{208}

eq10$.

Answer:

1. $(0,0)$
2. $4$
3. $x^{2}+y^{2}=16$
4. A. $(0,3)$