Question

If \( m_1 \) and \( m_2 \) are the slopes of the direct common tangents drawn to the circles \[ x^2 + y^2 - 2x - 8y + 8 = 0 \quad \text{and} \quad x^2 + y^2 - 8x + 15 = 0 \] then \( m_1 + m_2 \) is:

• A. \( \frac{-24}{5} \)
• B. \( \frac{12}{5} \)
• C. \( \frac{24}{5} \)
• D. \( \frac{-12}{5} \)

Hint:

For direct common tangents, compute centers and radii first, then use the tangent slope conditions.

Answer 1

The Correct Option is A

Let the equations of the circles be:
\(C_1: x^2 + y^2 - 2x - 8y + 8 = 0\)
\(C_2: x^2 + y^2 - 8x + 15 = 0\)
The center and radius of \(C_1\) are \(C_1(1, 4)\) and \(r_1 = \sqrt{1^2 + 4^2 - 8} = \sqrt{1 + 16 - 8} = \sqrt{9} = 3\).
The center and radius of \(C_2\) are \(C_2(4, 0)\) and \(r_2 = \sqrt{4^2 + 0^2 - 15} = \sqrt{16 - 15} = \sqrt{1} = 1\).
Let the equation of the direct common tangent be \(y = mx + c\).
For \(C_1\), the perpendicular distance from \(C_1\) to the tangent is \(r_1\).
\[\frac{|4 - m - c|}{\sqrt{1 + m^2}} = 3\] \[|4 - m - c| = 3\sqrt{1 + m^2}\] For \(C_2\), the perpendicular distance from \(C_2\) to the tangent is \(r_2\).
\[\frac{|0 - 4m - c|}{\sqrt{1 + m^2}} = 1\] \[|4m + c| = \sqrt{1 + m^2}\] \[c = -4m \pm \sqrt{1 + m^2}\] Substituting \(c\) into the first equation:
\[|4 - m - (-4m \pm \sqrt{1 + m^2})| = 3\sqrt{1 + m^2}\] \[|4 + 3m \mp \sqrt{1 + m^2}| = 3\sqrt{1 + m^2}\] Case 1: \(4 + 3m - \sqrt{1 + m^2} = 3\sqrt{1 + m^2}\) \(4 + 3m = 4\sqrt{1 + m^2}\)
\((4 + 3m)^2 = 16(1 + m^2)\)
\(16 + 24m + 9m^2 = 16 + 16m^2\)
\(7m^2 - 24m = 0\)
\(m(7m - 24) = 0\)
\(m = 0\) or \(m = \frac{24}{7}\)
Case 2: \(4 + 3m + \sqrt{1 + m^2} = 3\sqrt{1 + m^2}\)
\(4 + 3m = 2\sqrt{1 + m^2}\)
\((4 + 3m)^2 = 4(1 + m^2)\)
\(16 + 24m + 9m^2 = 4 + 4m^2\)
\(5m^2 + 24m + 12 = 0\)
Let \(m_3\) and \(m_4\) be the roots.
\(m_3 + m_4 = -\frac{24}{5}\)
\(m_3m_4 = \frac{12}{5}\)
The roots from Case 1 are \(m_1 = 0\) and \(m_2 = \frac{24}{7}\).
The roots from Case 2 are \(m_3\) and \(m_4\).
We need to find the slopes of the direct common tangents.
The sum of the slopes of the direct common tangents is \(m_1 + m_2\).
The sum of the roots of \(5m^2 + 24m + 12 = 0\) is \(-\frac{24}{5}\).
Since the slopes of the direct common tangents are the roots of the quadratic equation \(5m^2 + 24m + 12 = 0\), the sum of the slopes is \(m_1 + m_2 = -\frac{24}{5}\). Final Answer: The final answer is $\boxed{(1)}$