Question

The values of a for which the point (a – 1, a + 1) lies in the larger segment of the circle x² + y² – x – y – 6 = 0 made by the chord whose equation is x + y – 2 = 0 is –

• A. – 1 <a< 1
• B. 1 <a <
• C. – <a< – 1
• D. a £ 0

Answer 1

Answer: (A)

– 1 <a< 1

Answer 2

Circle S = x² + y² – x – y – 6 = 0

centre at (1/2, 1/2)

P(a – 1, a + 1) must lie inside the circle

So (a – 1)² + (a + 1)² – (a – 1) – (a + 1) – 6 < 0

a² – a – 2 < 0 Ž (a – 2) (a + 1) < 0– 1 < a < 2 … (1)

Also P and C must lie on the same side of the line

L = x + y – 2 = 0

L(1/2, 1/2) and L(a – 1, a + 1) must have same sign L$\left( \frac{1}{2},\frac{1}{2} \right)$ = $\frac{1}{2}$ + $\frac{1}{2}$ = – 1 < 0

L(a – 1, a + 1) = a – 1 + a + 1 – 2 < 0

a < 1 …(2)

By both the inequalities – 1 < a < 1.