Question

Let the centre of a circle C be (α, β) and its radius r < 8. Let 3x + 4y = 24 and 3x – 4y = 32 be two tangents and 4x + 3y =1 be a normal to C. Then (α – β + r) is equal to

• A. 5
• B. 6
• C. 7
• D. 9

Answer 1

The Correct Option is C

The centre of the circle lies on the normal: \[ 4\alpha + 3\beta = 1 \quad \cdots (1) \] The distance of the centre \( (\alpha, \beta) \) from the tangents \( L_1 \) and \( L_2 \) is equal: \[ \text{Distance of } (\alpha, \beta) \text{ from } L_1 = \text{Distance of } (\alpha, \beta) \text{ from } L_2. \] From the given equations of the tangents: \[ 3\alpha + 4\beta - 24 = -3\alpha - 4\beta + 32 \quad \Rightarrow \quad 6\alpha = 56 \quad \Rightarrow \quad \alpha = \frac{28}{3}, \quad \beta = \frac{-109}{3}. \] Now, the distance of \( (\alpha, \beta) \) from the line \( 3x + 4y = 24 \) is: \[ r = \sqrt{\left( \frac{28}{3} - 4 \right)^2 + \left( \frac{-109}{3} + 5 \right)^2} \quad (\text{reject as it is greater than 8}). \] For \( 3\alpha + 4\beta - 24 = -3\alpha - 4\beta + 32 \) we find: \[ \alpha = 1, \quad \beta = -1. \] Then: \[ \gamma = \sqrt{(4 - 1)^2 + (-5 + 1)^2} = 5. \] Therefore, \( \alpha - \beta + r = 7 \).