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: Understanding the Concept
We need to find the slopes of the two tangent lines that can be drawn from an external point to a circle. The condition for a line to be tangent to a circle is that the perpendicular distance from the center of the circle to the line is equal to the circle's radius. This condition will lead to a quadratic equation in the slope, $m$.
Step 2: Key Formula or Approach
1. Write the general equation of a line passing through the point $(-1,-2)$ with slope $m$: $y - (-2) = m(x - (-1)) \implies mx - y + (m-2) = 0$. 2. Identify the center and radius of the given circle. 3. Use the distance formula from the center to the tangent line and set it equal to the radius. 4. Solve the resulting quadratic equation in $m$. The roots will be $m_1$ and $m_2$. 5. Use the relationship between roots and coefficients ($m_1+m_2$, $m_1m_2$) to find the value of $|m_1-m_2|$ using the identity $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2}$. 6. Calculate the final expression $\sqrt{3}|m_1-m_2|$.
Step 3: Detailed Explanation
1. Equation of the tangent line: A line through $(-1,-2)$ with slope $m$ is $y+2=m(x+1)$, or $mx-y+m-2=0$. 2. Circle properties: The circle is $(x-3)^2+(y-4)^2=4$. The center is $(h,k)=(3,4)$. The radius is $r=\sqrt{4}=2$. 3. Apply the tangency condition: Distance from center $(3,4)$ to line $mx-y+m-2=0$ is equal to radius 2. \[ \frac{|m(3) - (4) + m-2|}{\sqrt{m^2 + (-1)^2}} = 2 \] \[ \frac{|4m-6|}{\sqrt{m^2+1}} = 2 \] \[ |4m-6| = 2\sqrt{m^2+1} \] 4. Solve for m: Square both sides: \[ (4m-6)^2 = (2\sqrt{m^2+1})^2 \] \[ 16m^2 - 48m + 36 = 4(m^2+1) \] \[ 16m^2 - 48m + 36 = 4m^2 + 4 \] \[ 12m^2 - 48m + 32 = 0 \] Divide by 4: \[ 3m^2 - 12m + 8 = 0 \] This quadratic equation has roots $m_1$ and $m_2$. 5. Find $|m_1-m_2|$: From Vieta's formulas for the quadratic $ax^2+bx+c=0$: Sum of roots: $m_1+m_2 = -(-12)/3 = 4$. Product of roots: $m_1m_2 = 8/3$. Now use the identity: \[ (m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2 \] \[ (m_1-m_2)^2 = (4)^2 - 4\left(\frac{8}{3}\right) = 16 - \frac{32}{3} = \frac{48-32}{3} = \frac{16}{3} \] \[ |m_1-m_2| = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}} \] 6. Calculate the final expression: \[ \sqrt{3}|m_1-m_2| = \sqrt{3} \times \frac{4}{\sqrt{3}} = 4 \] Step 4: Final Answer
The value of the expression $\sqrt{3}|m_1-m_2|$ is 4.