Given circle x² + y² + 4x + 6y - 5 = 0

Differentiate w.r.t x, we get

\(2x + 2y\frac{dy}{dx} + 4 + 6\frac {dy}{dx} = 0\)

⇒ \((2y + 6) \frac{dy}{dx} = - 2x - 4 \)

⇒ \(\frac{dy}{dx} = \frac{-2x - 4}{2y + 6} = \frac{-x -2}{y + 3}\)

Slope of line x + y = -13 is -1.

\(\therefore\) For point where slope of circle is -1, we have

\(\frac {-x_1 - 2}{y_1 + 3} = -1\)

⇒ \(-x_1 - 2 = -y_1 - 3\)

⇒ \(1 -x_1 = -y_1\)

⇒ \(y_1 = x_1 - 1\)

\(\therefore\) From equation of circle, we get

\({x_1}^2 + (x_1 - 2)^2 + 4x_1 + 6(x_1 - 1) - 5 = 0\)

⇒ \(2{x_1}^2 - 2x_1 + 1+ 4x_1 + 6x_1 - 6-5 =0\)

⇒ \(2{x_1}^2 + 8x_1 - 10 = 0\)

⇒ \({x_1}^2 + 4x_1 - 5 = 0\)

⇒ \((x_1 + 5)(x_1 - 1) = 0\)

⇒ \(x_1 = -5 \;or\; x_1 = 1\)

\(\therefore x_1 = 5\)

\(\therefore\) \(y_1 = -5 - 1 = -6\)

\(\therefore\) Point which on circle which is closest to line is (-5, -6).

The shortest distance between line & circle 

= Distance of point (-5, -6) from line x + y = -13

= \(\left|\frac{-5 + (-6) + 13}{\sqrt{1 + 1}}\right| = \frac 2{\sqrt 2} =\sqrt 2 \, units\)

Distance between (-5, -6) & (-6, -7)

\(= \sqrt{(-6 + 5)^2 + (-7 + 6)^2} = \sqrt{1 + 1} = \sqrt 2 \, units\)

& (-6, -7) lies on line x + y = -13.

\(\therefore\) Point on line x + y +13 = 0 which is closet to given circle is (-6, - 7).