Question

what is the general form of the equation of a circle with its Center at -2, 1 and passing through -4, 1

Answer 1

The given center is (-2,1), which means h = -2 and k = 1. We also know that the circle passes through (-4,1), which means x = -4 and y = 1.

Let's use the standard form first

`(x-h)^2+(y-k)^2=r^2`

Let's replace the given information to find r

`\begin{gathered} (-4-(-2))^2+(1-1)^2=r^2 \\ r=\sqrt[]{(-4+2)^2}=\sqrt[]{(-2)^2} \\ r=\sqrt[]{4}=\pm2 \end{gathered}`

Once we have r, we can look for the general form

`(x-(-2))^2+(y-1)^2=4`

Then, we solve the binomials

`\begin{gathered} (x+2)^2+(y-1)^2=4 \\ x^2+4x+4+y^2-2y+1=4 \end{gathered}`

At last, we organize all the terms and the left side.

`x^2+y^2+4x-2y+4+1-4=0`

Hence, the general form is

`x^2+y^2+4x-2y+1=0`