Question

The center of a circle is located at (3,8) and the circle has a radius that is 5 units long. What’s is the general form of the equation for the circle

Answer 1

Answer:

$\huge\boxed{x^2+y^2-6x-16y+48=0}$

Step-by-step explanation:

The standard form of the equation of a circle:

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

(h, k) - center of a circle

r - radius

We have the center (3, 8) → h = 3, k = 8

and the radius r = 5.

Substitute:

$(x-3)^2+(y-8)^2=5^2\\\\(x-3)^2+(y-8)^2=25$

The general form of the equation of a circle:

$x^2+y^2+Cx+Dy+E=0$

use $(a-b)^2=a^2+2ab+b^2$

$(x-3)^2+(y-8)^2=25\\\\x^2-(2)(x)(3)+3^2+y^2-(2)(y)(8)+8^2=25\\\\x^2-6x+9+y^2-16y+64=25\qquad\text{subtract 25 from both sides}\\\\x^2+y^2-6x-16y+9+64-25=0\\\\x^2+y^2-6x-16y+48=0$