Question

Let \( \theta \) be the angle between the circles \( S = x^2 + y^2 + 2x - 2y + c = 0 \) and \( S' = x^2 + y^2 - 6x - 8y + 9 = 0 \). If \( c \) is an integer and \( \cos\theta = \dfrac{5}{16} \), then the radius of the circle \( S = 0 \) is:

• A. 2
• B. 4
• C. 3
• D. 1

Hint:

The angle between two circles depends on their radii and the distance between centers. Use the formula involving \( \cos\theta \) to find unknowns when the angle is given.

Answer 1

The Correct Option is A

Step 1: Use the formula for angle between two circles 
Given two circles: 
\[ S = x^2 + y^2 + 2x - 2y + c = 0 \Rightarrow \text{center } C = (-1, 1), \text{ radius } r = \sqrt{1^2 + (-1)^2 - c} = \sqrt{2 - c} \] \[ S' = x^2 + y^2 - 6x - 8y + 9 = 0 \Rightarrow \text{center } C' = (3, 4), \text{ radius } r' = \sqrt{(-3)^2 + (-4)^2 - 9} = \sqrt{25 - 9} = 4 \] Step 2: Use formula for angle between two circles 
Let \( d = \text{distance between centers} = \sqrt{(3 + 1)^2 + (4 - 1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) 
Formula: 
\[ \cos\theta = \frac{r^2 + r'^2 - d^2}{2rr'} \] Substitute known values: 
\[ \cos\theta = \frac{(2 - c) + 16 - 25}{2 \cdot \sqrt{2 - c} \cdot 4} = \frac{-9 + (2 - c)}{8\sqrt{2 - c}} = \frac{-7 - c}{8\sqrt{2 - c}} = \frac{5}{16} \] Solve: 
\[ \frac{-7 - c}{\sqrt{2 - c}} = \frac{5}{2} \Rightarrow 2(-7 - c) = 5\sqrt{2 - c} \tag{1} \]