Question

The standard form of the equation of a circle is (x?4)2+(y?2)2=9. What is the general form of the equation? X2+y2+8x+4y+11=0 x2+y2+8x+4y?29=0 x2+y2?8x?4y?29=0 x2+y2?8x?4y+11=0

Answer 1

Answer:

$x^2+y^2-8x-4y+11=0$

Step-by-step explanation:

We want to find the equation of the circle: $(x-4)^2+(y-2)^2=9$ in general form.

We need to expand the parenthesis to obtain: $x^2-8x+16+y^2-4y+4=9$

This implies that:

$x^2+y^2-8x-4y+20=9$

We add -9 to both sides of the equattion to get:

$x^2+y^2-8x-4y+20-9=0$

Simplify the constant terms to get:

$x^2+y^2-8x-4y+11=0$