TS EAMCET · Maths · Circle
The slope of a common tangent to the circles \(x^2+y^2=16\) and \((x-9)^2+y^2=16\) is
- A \(\frac{8}{\sqrt{13}}\)
- B \(\frac{4}{\sqrt{13}}\)
- C \(\frac{\sqrt{17}}{8}\)
- D \(\frac{8}{\sqrt{17}}\)
Answer & Solution
Correct Answer
(D) \(\frac{8}{\sqrt{17}}\)
Step-by-step Solution
Detailed explanation
\(C_1(0,0), r_1=4\) \(C_2(9,0), r_2=4\) Tangent: \(mx-y+c=0\) \(\frac{|c|}{\sqrt{m^2+1}}=4 \Rightarrow c^2=16(m^2+1)\) \(\frac{|9m+c|}{\sqrt{m^2+1}}=4 \Rightarrow (9m+c)^2=16(m^2+1)\) \((9m+c)^2=c^2 \Rightarrow 9m+c=\pm c\) \(9m+c=c \Rightarrow m=0\) (Horizontal tangents)…
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