Question

25 POINTS AVAILABLE

1.
Identify the center and the radius of a circle with equation (x + 3)² + y² = 9
Format your answer as: (x,y), r
2.
Write the equation of a circle with center at (-4,3) and r = 6.
3.
Write the equation of a circle with the endpoints of the diameter at (-1, 6) and (5, -4). Watch the signs!

Answer 1

`\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(√(34))^2}`

Explanation:

The equation of a circle in standard form:

`(x-h)^2+(y-k)^2=r^2`

(h, k) - center

r - radius

1. We have the equation:

`(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2`

2. We have the center (-4, 3) and the radius r = 6. Substitute:

`(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36`

3. We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

`\left((x_1+x_2)/(2);\ (y_1+y_2)/(2)\right)`

Substitute:

`h=(-1+5)/(2)=(4)/(2)=2\\\\k=(6+(-4))/(2)=(2)/(2)=1`

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Substitute the coordinates of the points (2, 1) and (5, -4):

`r=√((5-2)^2+(-4-1)^2)=√(3^2+(-5)^2)=√(9+25)=√(34)`

Finally we have:

`(x-2)^2+(y-1)^2=(√(34))^2`