Question

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

• A. $12x+5y+19=0$
• B. $5x+12y+19=0$
• C. $5x-12y+19=0$
• D. $12x-5y+19=0$

Answer 1

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

Given circles,

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

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