If m_1, m_2 are the slopes of the tangents dr | vedclass

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(D) The equation of a line passing through $(-1, -2)$ with slope $m$ is $y + 2 = m(x + 1)$,which simplifies to $mx - y + (m - 2) = 0$. Since this line

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$4$

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(D) The equation of a line passing through $(-1, -2)$ with slope $m$ is $y + 2 = m(x + 1)$,which simplifies to $mx - y + (m - 2) = 0$. Since this line is tangent to the circle $(x-3)^2 + (y-4)^2 = 4$ with center $(3, 4)$ and radius $r = 2$,the perpendicular distance from the center to the line must equal the radius: $\frac{|m(3) - 4 + m - 2|}{\sqrt{m^2 + (-1)^2}} = 2$ $\frac{|4m - 6|}{\sqrt{m^2 + 1}} = 2$ $|2m - 3| = \sqrt{m^2 + 1}$ Squaring both sides: $(2m - 3)^2 = m^2 + 1$ $4m^2 - 12m + 9 = m^2 + 1$ $3m^2 - 12m + 8 = 0$ Let $m_1, m_2$ be the roots of this quadratic equation. Then $m_1 + m_2 = \frac{12}{3} = 4$ and $m_1 m_2 = \frac{8}{3}$. We know $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2} = \sqrt{4^2 - 4(\frac{8}{3})} = \sqrt{16 - \frac{32}{3}} = \sqrt{\frac{48 - 32}{3}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$. Therefore,$\sqrt{3}|m_1 - m_2| = \sqrt{3} 	imes \frac{4}{\sqrt{3}} = 4$.

If $m_1, m_2$ are the slopes of the tangents drawn through the point $(-1, -2)$ to the circle $(x-3)^2 + (y-4)^2 = 4$,then $\sqrt{3}|m_1 - m_2| = $

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