Question

The values of \(\alpha\) for which the point \((\alpha - 1, \alpha + 1)\) lies in the larger segment of the circle \(x^2 + y^2 - x - y - 6 = 0\) made by the chord whose equation is \(x + y - 2 = 0\) is

• A. \(-1<\alpha<1\)
• B. \(1<\alpha<\infty\)
• C. \(-\infty<\alpha<-1\)
• D. \(\alpha \leq 0\)

Hint:

Check both: inside circle and same side of chord as center.

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \text{Circle: } \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{13}{2} \]
Step 2: Calculation / Simplification}
Center \(C(1/2, 1/2)\). Chord: \(x+y-2=0\)
Point must lie inside circle: \((\alpha-1)^2 + (\alpha+1)^2 - (\alpha-1) - (\alpha+1) - 6<0\)
\(\alpha^2 - 2\alpha + 1 + \alpha^2 + 2\alpha + 1 - 2\alpha - 6<0\)
\(2\alpha^2 - 2\alpha - 4<0 \Rightarrow \alpha^2 - \alpha - 2<0\)
\((\alpha-2)(\alpha+1)<0 \Rightarrow -1<\alpha<2\)
Point and center must be on same side of chord:
\(C\): \(1/2+1/2-2 = -1<0\)
Point: \((\alpha-1)+(\alpha+1)-2 = 2\alpha-2<0 \Rightarrow \alpha<1\)
Intersection: \(-1<\alpha<1\)
Step 3: Final Answer
\[ -1<\alpha<1 \]