Question

A certain circle can be represented by the following equation. x^2+y^2+8x-16y+31=0x 2 +y 2 +8x−16y+31=0x, squared, plus, y, squared, plus, 8, x, minus, 16, y, plus, 31, equals, 0 What is the center of this circle ? ((left parenthesis ,,comma ))right parenthesis What is the radius of this circle ? units

Answer 1

center: (4,-4)

Radius: 9

Explanation:

KHAN ACADEMY

Answer 2

center =
`(-4,8)`

Radius = 7 units

Explanation:

Given: Equation of circle is
`x^2+y^2+8x-16y+31=0`

To find: Radius and center of the circle

Solution:

Equation of circle is
`(x-a)^2+(y-b)^2=r^2`

Here,
`(a,b)` is the center and r is the radius.

`x^2+y^2+8x-16y+31=0\\\left [ x^2+2(4)x+4^2 \right ]+\left [ y^2-2(8)y+8^2 \right ]+31=4^2+8^2`

Use formula
`(u+v)^2=u^2+v^2+2uv`

`(x+4)^2+(y-8)^2=16+64-31\\(x+4)^2+(y-8)^2=49=7^2`

On comparing this equation with equation of circle,

center =
`(-4,8)`

Radius = 7 units