114k views
2 votes
Z^2+ y^2 + 8x – 16y + 31 = 0

What is the center of this circle ?
What is the radius of this circle ?

User VeenarM
by
7.7k points

1 Answer

3 votes

Answer:

(- 4, - 8 ), radius = 7

Explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + y² + 8x - 16y + 31 = 0

Collect the x- terms, collect the y- terms together and subtract 31 from both sides.

x² + 8x + y² - 16y = - 31

Use the method of completing the square on the x and y terms

add ( half the coefficient of the x/ y term )² to both sides

x² + 2(4)x + 16 + y² + 2(- 8)y + 64 = - 31 + 16 + 64

(x + 4)² + (y - 8)² = 49 ← in standard form

with centre = (- 4, 8 ) and r =
√(49) = 7

User Yttrill
by
7.9k points