Question

in exercises 9 and 10, use the given information to write the standard equation of the circle.

1. the center is (0,0) and a point on the circle is (3, -4)
2. the center is (3, -2) and a point on the circle is (23, 19)
3. prove or disprove that the point (-3, 3) lies on the circle centered at the origin with radius 4.
4. the circumference of a circle is 10π with a center of (-2, -5). solve for the radius using the circumference formula c = 2πr and then write the equation of the circle.

Explanation:

Exercise 9

Step1: Recall circle standard form

Standard equation: $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is center, $r$ is radius.

Step2: Calculate radius squared

Center $(0,0)$, point $(3,-4)$. $r^2=(3-0)^2+(-4-0)^2=9+16=25$

Step3: Substitute into equation

Substitute $h=0,k=0,r^2=25$: $x^2+y^2=25$

Exercise 10

Step1: Recall circle standard form

Standard equation: $(x-h)^2+(y-k)^2=r^2$

Step2: Calculate radius squared

Center $(3,-2)$, point $(23,19)$. $r^2=(23-3)^2+(19-(-2))^2=20^2+21^2=400+441=841$

Step3: Substitute into equation

Substitute $h=3,k=-2,r^2=841$: $(x-3)^2+(y+2)^2=841$

Exercise 11

Step1: Recall circle standard form

Circle at origin, $r=4$: $x^2+y^2=16$

Step2: Test point $(-3,3)$

$(-3)^2+3^2=9+9=18
eq16$

Step3: Conclude result

Since $18
eq16$, point is not on the circle.

Exercise 12

Step1: Solve for radius $r$

Use $C=2\pi r$, $C=10\pi$. $10\pi=2\pi r \implies r=5$

Step2: Recall circle standard form

Standard equation: $(x-h)^2+(y-k)^2=r^2$

Step3: Substitute values

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

Answer:

1. $x^2+y^2=25$
2. $(x-3)^2+(y+2)^2=841$
3. The point $(-3,3)$ does not lie on the circle, since $(-3)^2+3^2=18

eq16$.

1. Radius $r=5$; circle equation: $(x+2)^2+(y+5)^2=25$