The equation of the cirele which passes through the point (1, 1) and touches the circle `x^2+y^2+4x-6y-3=0` at the point (2, 3) on it is

Explanation

To find the equation of the circle that passes through the point (1, 1) and touches the circle given by the equation \(x^2 + y^2 + 4x - 6y - 3 = 0\) at the point (2, 3), we will follow these steps:

### Step 1: Rewrite the given circle's equation in standard form
The equation of the circle is given as:
\[ x^2 + y^2 + 4x - 6y - 3 = 0 \]
We can complete the square for both \(x\) and \(y\).
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