Question

Write the standard form of the equation of the circle with the given characteristics.

Endpoints of a diameter: (2,3), (-8,-9)

Answer 1

The standard equation of a circle with diameter endpoints at (2,3) and (-8,-9) is (x + 3)^2 + (y + 3)^2 = 61, where the center of the circle is located at (-3, -3) with a radius of √[61].

Step-by-step explanation:

To write the standard equation of the circle, we need to find the center and the radius of the circle. The center of the circle is the midpoint of the diameter. Given the endpoints of the diameter (2,3) and (-8,-9), we can find the midpoint by calculating the average of the x-coordinates and the y-coordinates separately.

The midpoint (center of the circle) (x,y) can be found using:

• x = (x1 + x2) / 2
• y = (y1 + y2) / 2

So the center (h,k) is:

• h = (2 - 8) / 2 = -3
• k = (3 - 9) / 2 = -3

The radius (r) can be found by using the distance formula between one endpoint and the center:

r = √[(x2 - h)^2 + (y2 - k)^2]

The calculated radius is:

• r = √[(-8 + 3)^2 + (-9 + 3)^2]
• r = √[25 + 36]
• r = √[61]

Therefore, the standard equation of the circle is:

(x + 3)^2 + (y + 3)^2 = 61

Answer 2

(x +3)^2 + (y +3)^2 = 61

Step-by-step explanation:

The standard form equation of a circle with center (h, k) and radius r is ...

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

So, to write the desired equation, we need to know the center of the circle and its radius.

The center is the midpoint of the diameter, so is the average of the endpoints:

((2, 3) +(-8, -9))/2 = ((2-8)/2, (3-9)/2) = (-6/2, -6/2) = (-3, -3)

The radius can be found using the distance formula to find the distance between the circle center and either of the diameter end points.

d = √((x2 -x1)^2 +(y2 -y1)^2)

r = √((-3-2)^2 +(-3-3)^2) = √(25+36) = √61

Then ...

• (h, k) = (-3, -3)
• r^2 = (√61)^2 = 61

so the desired equation is ...

(x +3)^2 + (y +3)^2 = 61