Question

If $m_1$ and $m_2$ are the slopes of tangents to the circle $x^2+y^2=4$ from the point $(3,2)$, then $m_1-m_2$ is equal to

• A. $\frac{5}{12}$
• B. $\frac{12}{5}$
• C. $\frac{3}{2}$
• D. $0$

Answer 1

The correct option is B $\frac{12}{5}$

Equation of pair of tangents is $S{S}_1=T^2$
$\Rightarrow (x^2+y^2-4)(9+4-4)=(3x+2y-4{)}^2$
$\Rightarrow 5y^2+16y-12xy+24x-52=0$
$\therefore m_1+m_2=\frac{-2h}{b}=\frac{12}{5}$
and $m_1{m}_2=0$
Now, $m_1-m_2=\sqrt{(m_1+m_2{)}^2-4m_1{m}_2}$
$=\sqrt{{\left(\frac{12}{5}\right)}^2-0}$
$=\frac{12}{5}$