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).