Question

The equation of the common tangent of the two touching circles, $y^2+x^2-6x-12y+37=0$ and $x^2+y^2-6y+7=0$ is:

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

Answer 1

The correct option is C

$x-y-5=0$

Explanation for the correct answer:

Finding the equation of tangent,

Given that the equation of the two circles are $y^2+x^2-6x-12y+37=0$ and $x^2+y^2-6y+7=0$

The equation for the tangent is:

$S_1-S_2=0$

By substituting the equations we get

$\begin{array}{rcl}{y}^2+x^2-6x-12y+37-(x^2+y^2-6y+7)& =& 0\\ -6x+6y+30& =& 0\\ 6x-6y-30& =& 0\\ x-y-5& =& 0\end{array}$

Hence, the correct answer is Option (C).