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Archy [21]
3 years ago
8

Help with algebra hw

Mathematics
2 answers:
AnnZ [28]3 years ago
6 0
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
Karo-lina-s [1.5K]3 years ago
5 0
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.
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