Find the equation of the normals to the circle `x^2+y^2-8x-2y+12=0` at the point whose ordinate is `-1`

Explanation

To find the equation of the normals to the circle \(x^2 + y^2 - 8x - 2y + 12 = 0\) at the point where the ordinate is \(-1\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. We can do this by completing the square.

The given equation is:
\[
x^2 + y^2 - 8x - 2y + 12 = 0
...