Find the equation of the circle whose centre is at the point `(4,5)` and which passes through the centre of the circle `x^2+y^2-6x+4y-12=0.`

Explanation

To find the equation of the circle with center at the point (4, 5) that passes through the center of the circle defined by the equation \(x^2 + y^2 - 6x + 4y - 12 = 0\), we will follow these steps:

### Step 1: Find the center of the given circle
The equation of the circle is given in the standard form \(x^2 + y^2 + Dx + Ey + F = 0\). We can rewrite the given equation \(x^2 + y^2 - 6x + 4y - 12 = 0\) in the form of completing the square.

1. Rearranging the equation:
\[
x^2 - 6x + y^2 + 4y - 12 = 0
...