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

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

Step2: Calculate radius squared

Use distance formula: $r^2=(3-0)^2+(-4-0)^2=9+16=25$

Step3: Substitute into standard form

Substitute $(h,k)=(0,0)$ and $r^2=25$:
$(x-0)^2+(y-0)^2=25$

Exercise 10

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, $r$ is radius.

Step2: Calculate radius squared

Use distance formula: $r^2=(23-3)^2+(19-(-2))^2=20^2+21^2=400+441=841$

Step3: Substitute into standard form

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

Exercise 11

Step1: Recall circle standard form

Circle centered at origin with radius 4: $x^2+y^2=16$

Step2: Substitute point into equation

Substitute $(-3,3)$: $(-3)^2+3^2=9+9=18$

Step3: Compare to radius squared

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

Exercise 12

Step1: Solve for radius $r$

Use $C=2\pi r$, substitute $C=10\pi$:
$10\pi=2\pi r \implies r=\frac{10\pi}{2\pi}=5$

Step2: Calculate radius squared

$r^2=5^2=25$

Step3: Substitute into standard form

Substitute $(h,k)=(-2,-5)$ and $r^2=25$:
$(x+2)^2+(y+5)^2=25$

Answer:

1. $\boldsymbol{x^2+y^2=25}$
2. $\boldsymbol{(x-3)^2+(y+2)^2=841}$
3. The point $(-3,3)$ does NOT lie on the circle centered at the origin with radius 4.
4. Radius: $\boldsymbol{5}$; Circle equation: $\boldsymbol{(x+2)^2+(y+5)^2=25}$