TS EAMCET · Maths · Circle
The line \(4 x-3 y+2=0\) intersects the circle \(x^2+y^2-2 x+6 y+c=0\) at two points A, B and \(\mathrm{AB}=8\). If \((1, \mathrm{k})\) is a point on the given circle and \(\mathrm{k}>0\), then \(\mathrm{k}=\)
- A 8
- B 4
- C 2
- D 1
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
Center of circle: \((1, -3)\). Radius of circle: \(R = \sqrt{(-1)^2 + (3)^2 - c} = \sqrt{10 - c}\). Distance from center \((1, -3)\) to line \(4x - 3y + 2 = 0\): \(d = \frac{|4(1) - 3(-3) + 2|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 + 9 + 2|}{\sqrt{16 + 9}} = \frac{15}{5} = 3\). Using…
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