what is the general form of the equation of the circle shown?
![what is the general form of the equation of the circle shown class=](https://us-static.z-dn.net/files/dd0/4c4d32bddb2ee0488352d29de388730c.png)
[tex](x-(-2))^2+(y-1)^2=3^2\\(x+2)^2+y^2-2y+1-9=0\\x^2+4x+4+y^2-2y-8=0\\x^2+y^2+4x-2y-4=0[/tex]
Answer:
x²+4x+y²−2y−4=0
Step-by-step explanation:
we can observe that,
1) the coordinates of the center of the circle (h,k) = (-2,1)
2) the radius, r = 3
using the standard form of the equation of circle:
(x−h)²+(y−k)²=r² (substitute h= -2, k=1 and r=3), we get
(x+2)²+(y−1)²=9 ----> this is in "Standard" form, to get "General" form, simply expand the parenthesis reduce to simplest terms.
(x+2)²+(y−1)²=9
[x² + (2)(x)(2) + 2²] + [y² + (2)(y)(-1) + (-1)²] = 9 (expand and reduce)
x²+4x+y²−2y−4=0 (answer)