AP EAMCET · Maths · Circle
The point on the circle \(x^2+y^2=4\) whose distance from the line \(4 x+3 y-12=0\) is \(4 / 5\) units is equal to
- A \(\left(\frac{12}{25}, \frac{36}{25}\right)\)
- B \((4,0)\)
- C \((2,0)\)
- D \(\left(\frac{-14}{25}, \frac{48}{25}\right)\)
Answer & Solution
Correct Answer
(C) \((2,0)\)
Step-by-step Solution
Detailed explanation
Let the point be \((h, k)\) So, \(\quad \frac{|4 h+3 k-12|}{5}=\frac{4}{5} \Rightarrow|4 h+3 k-12|=4\) \(\Rightarrow \quad(4 h+3 k=16)\) or \((4 h+3 k=8)\) \((h, k)\) lies on circle so…
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