Question

Help with algebra hw

Answer 1

It's a equation of the circle.

The standard form equation of a circle:


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

where (h; k) - the coordinates of the center of the circle
r - the radius

We have:


`x^2+y^2-8x+6y+5=0`

Use:
`(a\pm b)^2=a^2\pm 2ab+b^2`



`x^2-8x+y^2+6y+5=0\\\\\underbrace{x^2-2\cdot x\cdot4+4^2}_((a-b)^2=a^2-2ab+b^2)-4^2+\underbrace{y^2+2\cdot y\cdot3+3^2}_((a+b)^2=a^2+2ab+b^2)-3^2+5=0\\\\(x-4)^2+(y+3)^2-16-9+5=0\\\\(x-4)^2+(y+3)^2-20=0\ \ \ |+20\\\\(x-4)^2+(y+3)^2=20\\\\(x-4)^2+(y+3)^2=(√(20))^2\\\\(x-4)^2+(y+3)^2=(2\sqrt5)^2`


The center of the circle (4; -3).
The length of radius 2√5.

Answer 2

x² + y² - 8x + 6y + 5 = 0
x² - 8x +16 + y² + 6y + 9 + 5 - 16 - 9 = 0
(x - 4)² + (y + 3)² - 20 = 0
(x - 4)² + (y + 3)² = 20

Circle with center (4, -3) and radius 2√5