Question

Question #e6f38

Answer 1

`x^2 + y^2 + 3x +2y = 0`

Explanation:

If a diameter of the circle has endpoints `(0,0)` and `(-3,-4)`, the center of the circle is in the midpoint of the diameter and has coordinates:

`(x_x,y_c) = ( ( -3 + 0)/2, (-4+0)/2) = (-3/2,-2)`

while the radius has length:

`r= 1/2 sqrt ( ( -3 - 0)^2+ (-4-0)^2) = 1/2 sqrt(9+16) = 5/2`

The general equation of the circle is:

`(x-x_c)^2+(y-y_c)^2 = r^2`

that is:

`(x+3/2)^2+(y+2)^2 = 25/4`

or:

`(2x+3)^2+4(y+2)^2 = 25`

`4x^2 +12x + 9 +4y^2 +8y +16 = 25`

`4x^2 + 4y^2 + 12x +8y = 0`

`x^2 + y^2 + 3x +2y = 0`

graph{x^2 + y^2 + 3x +2y = 0 [-10, 10, -5, 5]}