AP EAMCET · Maths · Circle
The radius of the circle having \(3 x-4 y+4=0\) and \(6 x-8 y-7=0\) as its tangents is
- A \(\frac{3}{2}\)
- B \(3\)
- C \(6\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
Equation of the given tangents are \(\begin{aligned} & E_1=3 x-4 y+4=0 \\ & E_2=6 x-8 y-7=0\end{aligned}\) \(\Rightarrow \quad 3 x-4 y-\frac{7}{2}=0\) Here, \(a=3, b=-4 c_1=4, c_2=\frac{-7}{2}\) Since, slopes of the given tangents are equal, i.e. \(\frac{3}{4}\). \(\therefore\)…
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