If m_{1} and m_{2} are the slopes of tangents | vedclass

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(B) The equation of the pair of tangents from a point $(x_{1}, y_{1})$ to the circle $S: x^{2}+y^{2}-r^{2}=0$ is given by $SS_{1}=T^{2}$.Here,$S = x^{

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$\frac{12}{5}$

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(B) The equation of the pair of tangents from a point $(x_{1}, y_{1})$ to the circle $S: x^{2}+y^{2}-r^{2}=0$ is given by $SS_{1}=T^{2}$.Here,$S = x^{2}+y^{2}-4$,$S_{1} = 3^{2}+2^{2}-4 = 9$,and $T = 3x+2y-4$.Substituting these,we get $(x^{2}+y^{2}-4)(9) = (3x+2y-4)^{2}$.Expanding the equation: $9x^{2}+9y^{2}-36 = 9x^{2}+4y^{2}+16+12xy-24x-16y$.Simplifying: $5y^{2}-12xy+24x+16y-52 = 0$.For a pair of lines $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,the slopes $m_{1}, m_{2}$ satisfy $bm^{2}+2hm+a=0$ (by dividing by $x^{2}$ and setting $m=y/x$).Here,$b=5$,$2h=-12$ (so $h=-6$),and $a=24$ (coefficient of $x^{2}$ is $0$,but we treat the quadratic in $y/x$).Actually,dividing $5y^{2}-12xy+24x+16y-52=0$ by $x^{2}$ gives $5m^{2}-12m+24/x+16m/x-52/x^{2}=0$. This approach is complex. Using the condition for tangents $y-2 = m(x-3) \Rightarrow mx-y+(2-3m)=0$.The distance from $(0,0)$ to the line is $r=2$: $\frac{|2-3m|}{\sqrt{m^{2}+1}} = 2$.Squaring both sides: $(2-3m)^{2} = 4(m^{2}+1)$.$4-12m+9m^{2} = 4m^{2}+4$.$5m^{2}-12m = 0$.$m(5m-12) = 0$.Thus,the slopes are $m_{1} = 0$ and $m_{2} = \frac{12}{5}$.Therefore,$|m_{1}-m_{2}| = |0 - \frac{12}{5}| = \frac{12}{5}$.

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

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