Question

Help with algebra hw

Answer 1

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

Answer 2

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


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


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