Question

Find the Slope of the transverse common tangent to the circles $x^2+y^2-4x+6y-12=0$ and $x^2+y^2+6x+18y+26=0$

• A. $\frac{-5}{12}$
• B. $\frac{5}{12}$
• C. $\frac{12}{5}$
• D. $\frac{-12}{5}$

Answer 1

The correct option is A

$\frac{-5}{12}$

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.

$c_1(2,3)$, $c_2(-3,-9)$

$r_1=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=5$

$r_2=\sqrt{g^2+f^2-c}=\sqrt{9+81-26}=8$

$c_1{c}_2=\sqrt{(2+3{)}^2}+(3+9{)}^2=\sqrt{25+144}=\sqrt{169}=13$

$c_1{c}_2=r_1+r_2$

Both the circle touch each other externally.

Transverse common tangent is perpendicular to the line segment c1c2 and it divides line segment in the ratio of r1 and r2.

Product of slope of the two perpendicular line = -1

Slope of the line segment $c_1{c}_2\left(m_1\right)=\frac{y_2-y_1}{x_2-x_1}=\frac{-9-3}{-3-2}=\frac{-12}{-5}=\frac{12}{5}$

Slope of transverse common tangent =$\frac{-1}{m_1}=\frac{-1}{\frac{12}{5}}=\frac{-5}{12}$

Correct option is A.