TS EAMCET · Maths · Circle
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 \(\sqrt{3}\left|m_1-m_2\right|=\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(m x - y + (m+2) - 4 = 0 \implies m x - y + m - 2 = 0\) \(\frac{|3m - 4 + m - 2|}{\sqrt{m^2 + (-1)^2}} = 2\) \((4m - 6)^2 = 4(m^2 + 1)\) \(16m^2 - 48m + 36 = 4m^2 + 4\) \(12m^2 - 48m + 32 = 0\) \(3m^2 - 12m + 8 = 0\)…
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