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

Question

Dr. Abubacker · Accepted Answer

(b)
Equation of pair of tangents is
$SS_{1} = T^{2}$
$\Rightarrow \ \left( x^{2} + y^{2} - 4 ight)(9 + 4 - 4) = (3x + 2y - 4)^{2}$
$\Rightarrow \ 5y^{2} + 16y - 12xy + 24x - 50 = 0$
$	herefore\ m_{1} + m_{2} = - \frac{2h}{b} = \frac{12}{5}$
and $m_{1}m_{2} = 0$
Now, $m_{1} - m_{2} = \sqrt{\left( m_{1} + m_{2} ight)^{2} - 4m_{1}m_{2}}$
$= \sqrt{\left( \frac{12}{5} ight)^{2} - 0} = \frac{12}{5}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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