Equation of a Circle
Using the following equations, find the center and radius of the circle:
- $(x-5)^2+(y-9)^2=81$: center: (5, 9), radius: 9
- $(x-2)^2+(y-4)^2=25$ center: (2, 4), radius: 5
- $(x+4)^2+(y-7)^2=49$ center: (-4, 7), radius: 7
- $(x+6)^2+(y+1)^2=16$ center: (-6, -1), radius: 4
Graph the following circles:
$(x+3)^2+(y+2)^2=4$
$(x-1)^2+(y-1)^2=9$
$(x+2)^2+(y+1)^2=16$
Test to see if the following points are on the circle with equation $(x-4)^2+(y-5)^2=81$:
- $(13,14)$ No $\eqalign{(13-4)^2+(14-5)^2=81\\9^2+9^2=81\\81+81=81\\162 \neq 81}$
- $(4,-4)$ Yes $\eqalign{(4-4)^2+(-4-5)^2=81\\0^2+(-9)^2=81\\0+81=81\\81=81 81}$
- $(2,2)$ No $\eqalign{(2-4)^2+(2-5)^2=81\\(-2)^2+(-3)^2=81\\4+9=81\\13 \neq 81}$
- $(0,5)$ No $\eqalign{(0-4)^2+(5-5)^2=81\\(-4)^2+0^2=81\\16+0=81\\16 \neq 81}$
Write the equations for the following circles:
$h=-2, k=3, r=2 \rightarrow (x+2)^2+(y-3)^2=4$
$h=0, k=-3, r=4 \rightarrow (x-0)^2+(y+3)^2=4$
$h=4, k=-2, r=1 \rightarrow (x-4)^2+(y+2)^2=1$
$h=0, k=1, r=5 \rightarrow (x-0)^2+(y-1)^2=25$$h=-4, k=-4, r=2 \rightarrow (x+4)^2+(y+4)^2=4$