Question

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 $\sqrt{3}|m_1-m_2|=$

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For problems involving the difference of roots ($|x_1-x_2|$) of a quadratic equation $ax^2+bx+c=0$, you can use the direct formula $|x_1-x_2| = \frac{\sqrt{D}}{a}$, where $D=b^2-4ac$ is the discriminant. For $3m^2-12m+8=0$, $D=(-12)^2-4(3)(8) = 144-96=48$. So $|m_1-m_2|=\frac{\sqrt{48}}{3}=\frac{4\sqrt{3}}{3}=\frac{4}{\sqrt{3}}$.

Answer 1

The Correct Option is D

Step 1: Tangent Condition: Line through \( (-1,-2) \) with slope \( m \): \( y+2 = m(x+1) \implies mx - y + (m-2) = 0 \). Circle Centre \( (3,4) \), Radius \( r=2 \). Distance from centre to tangent equals radius: \[ \frac{|m(3) - 4 + m - 2|}{\sqrt{m^2+1}} = 2 \] \[ \frac{|4m - 6|}{\sqrt{m^2+1}} = 2 \] \[ |2m - 3| = \sqrt{m^2+1} \]
Step 2: Solve for m: Square both sides: \[ (2m-3)^2 = m^2+1 \] \[ 4m^2 - 12m + 9 = m^2 + 1 \] \[ 3m^2 - 12m + 8 = 0 \]
Step 3: Calculate \( |m_1-m_2| \): Difference of roots for \( am^2+bm+c=0 \) is \( \frac{\sqrt{b^2-4ac}}{|a|} \). \[ |m_1-m_2| = \frac{\sqrt{(-12)^2 - 4(3)(8)}}{3} = \frac{\sqrt{144-96}}{3} = \frac{\sqrt{48}}{3} = \frac{4\sqrt{3}}{3} \]
Step 4: Final Value: \[ \sqrt{3}|m_1-m_2| = \sqrt{3} \times \frac{4\sqrt{3}}{3} = \frac{4 \times 3}{3} = 4 \]