Hello!
The answer is:
The general form of the circle is:
![x^(2) -2x+y^(2) +4y=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vyu6v2niajdgtdxzsiuk7ofwjr34nuqt1x.png)
Why?
We are given the standard form of a circle, so, to calculate the general form, we need to solve notable products.
We must remember the way to solve the notable product:
![(a+b)^(2)=a^(2)+2ab+b^(2) \\\\(a-b)^(2)=a^(2)-2ab+b^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/l389g97igxm65052tepo2d1bdpd8qb6pjm.png)
So,
We are given the equation:
![(x-1)^(2)+(y+2)^(2)=5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/emgyniimbt097b87n1505mp4ecmd27yr4e.png)
Then, solving the notable products, and adding/subtracting like terms, we have:
![x^(2) -2*1x+1+y^(2)+2*2y+4=5\\x^(2) -2x+1+y^(2) +4y+4=5\\\\x^(2) -2x+y^(2) +4y+1+4-5=0\\\\x^(2) -2x+y^(2) +4y=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/q0juxrqy686yr9ghjfwfoi06r0y5mn6gir.png)
Hence, have that the general form of the circle is:
![x^(2) -2x+y^(2) +4y=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vyu6v2niajdgtdxzsiuk7ofwjr34nuqt1x.png)
Have a nice day!