Question

What is the equation of the circle in general form?

Answer 1

The Cartesian form is:

`(x - (-1))^2+(y-1)^2= 3^2`

Explanation:

The general Cartesian form for the equation of a circle is:

`(x - h)^2+(y-k)^2= r^2" [1]"`

Where `(x,y)` is any point on the circle, `(h,k)` is the center point of the circle, and `r` is the radius.

We can see that the center point is `(-1,1)`

Substitute this into equation [1]:

`(x - (-1))^2+(y-1)^2= r^2" [2]"`

We can see that the circle intersects the point `(-1, 4)`; we can substitute this into equation [2] to find the radius:

`(-1 - (-1))^2+(4-1)^2= r^2`

`0^2 + 3^2 = r^2`

`r = 3`

Substitute the radius into equation [2]:

`(x - (-1))^2+(y-1)^2= 3^2" [3]"`

Note: One can change `(x - (-1))^2` into `(x +1)^2` but I do not recommend it.

Answer 2

`x^2+y^2+2x-2y-7=0`

Explanation:

`"the "color(blue)"general equation of a circle"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(x^2+y^2+2gx+2fy+c=0)color(white)(2/2)|)))`

`"with centre "=(-g,-f)" and radius " sqrt(g^2+f^2-c)`

`"here centre "=(-1,1)`

`rArrg=1" and "f=-1`

`"here radius "=3`

`rArrg^2+f^2-c=9`

`rArrc=1^2+(-1)^2-9=-7`

`rArrx^2+y^2+2x-2y-7=0larrcolor(red)" in general form"`