Question

The circle \(x^2 + y^2 - 8x - 12y + \alpha = 0\) lies in the first quadrant without touching the coordinate axes. If \((6, 6)\) is an interior point to the circle, then the range of \(\alpha\) is:

• A. \(4<\alpha<6\)
• B. \(6<\alpha<16\)
• C. \(16<\alpha<48\)
• D. \(36<\alpha<48\)

Hint:

When determining if a point is inside a circle defined by an equation, substitute the point into the equation and compare with the radius squared.

Answer 1

The Correct Option is D

Step 1: Complete the square to find the center and radius of the circle. For the equation: \[ x^2 - 8x + y^2 - 12y + \alpha = 0, \] complete the square: \[ (x-4)^2 - 16 + (y-6)^2 - 36 + \alpha = 0, \] which simplifies to: \[ (x-4)^2 + (y-6)^2 = 52 - \alpha. \] Step 2: Analyze the position of point \((6,6)\) relative to the circle. Substitute \((6,6)\) into the circle's equation: \[ (6-4)^2 + (6-6)^2 = 4. \] So, the radius squared is \(52 - \alpha\), and we need \(4<52 - \alpha\) to ensure \((6,6)\) is inside the circle. 
Step 3: Ensure the circle does not touch the axes. This means both \(x\) and \(y\) intercepts must be positive, i.e., the radius must be less than the distance from the center to either axis. 
Step 4: Compute the valid range for \(\alpha\): \[ 4<52 - \alpha \quad \Rightarrow \quad \alpha<48, \] and since the center is at \((4,6)\), the smallest radius to not touch the axes is the distance to the x-axis (4 units), hence: \[ r>4 \quad \Rightarrow \quad 52 - \alpha>16 \quad \Rightarrow \quad \alpha<36. \] The correct conditions are \(\alpha>36\) and \(\alpha<48\).