TS EAMCET · Maths · Circle
If \(\alpha \neq-4\) and \((2, \alpha)\) is the mid-point of a chord of the circle \(x^2+y^2-4 x+8 y+6=0\), then the values of the \(y\)-intercept of the chord lie in the interval
- A \((-4-\sqrt{14},-4+\sqrt{14})\)
- B \((-4,4)\)
- C \((4-\sqrt{14}, 4+\sqrt{14})\)
- D \((-2,2)\)
Answer & Solution
Correct Answer
(A) \((-4-\sqrt{14},-4+\sqrt{14})\)
Step-by-step Solution
Detailed explanation
We have, \((2, \alpha)\) is mid-point of chord of circle \(x^2+y^2-4 x+8 y+6=0\) \(\because(2, \alpha)\) inside the circle \(\therefore \quad 4+\alpha^2-8+8 \alpha+6 \leq 0\) \(\alpha^2+8 \alpha+2 \leq 0\) \(\alpha=\frac{-8 \pm \sqrt{64-8}}{2}\)…
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