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AP EAMCET · Maths · Circle

The shortest distance from the line \(3 x+4 y=25\) to the circle \(x^2+y^2-6 x+8 y=0\) is

  1. A \(\frac{9}{5}\)
  2. B \(\frac{7}{5}\)
  3. C \(\frac{8}{5}\)
  4. D \(\frac{13}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{7}{5}\)

Step-by-step Solution

Detailed explanation

We have a line \(3 x+4 y=25\) and a circle \(x^2+y^2-6 x+8 y=0\) for shortest distance. We draw a perpendicular from \(\mathrm{C}=\) \((3,-4)\) to the straight line. \(\Rightarrow \mathrm{d}=\frac{|9-16-25|}{\sqrt{9+16}}=\frac{32}{5}\) hence shortcut distance…