The equation of the common tangent to the circles $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ at their point of contact.

12x + 5y + 19 = 0

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5x + 12y + 19 = 0

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5x - 12y + 19 = 0

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12x - 5y + 19 = 0

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Solution

The correct option is B
5x + 12y + 19 = 0

Given circles,

$x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ - - - - - - (1)

$x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ - - - - - - (2)

Let the circle be $c_{1} & c_{2}$ and radii $r_{1} & r_{2}$ of circle (1) and (2) respectively.

$c_{1} (2, 3)$ , $c_{2} (- 3, - 9)$

$r_{1} = \sqrt{g^{2} + f^{2} - c} = \sqrt{4 + 9 + 12} = 5$

$r_{2} = \sqrt{g^{2} + f^{2} - c} = \sqrt{9 + 81 - 26} = 8$

$c_{1} c_{2} = \sqrt{(2 + 3)^{2}} + (3 + 9)^{2} = \sqrt{25 + 144} = \sqrt{169} = 13$

$c_{1} c_{2} = r_{1} + r_{2}$

So,both the circle touch each other externally.

Here, tangent at the point of contact is a transverse common tangent point of intersection p divides the line joining center of circles internally in the ratio of their radii. $r_{1} : r_{2} = 5 : 8$

Transverse common tangent is perpendicular to the line segment $c_{1} c_{2} .$

Product of slope of the two perpendicular line = -1

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The equation of the common tangent to the circles x2 + y2 − 4x + 6y − 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0 at their point of contact.

The equation of the common tangent to the circles $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ at their point of contact.