Circles
Position of a Point
Grade 11
Question:
<p>In each of the following one or more options are correct. Choose the correct option(s).</p><p>(b) If (2, 5) is an interior point of the circle \(x^2 + y^2 - 8x - 12y + p = 0\) and the circle neither cuts nor touches any one of the axes of coordinates then</p>
<p>A. \(p \in (36, 47)\)</p>
<p>B. \(p \in (16, 47)\)</p>
<p>C. \((16, 36)\)</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: A point is interior to a circle if substituting its coordinates makes the LHS of the circle equation negative. For the circle to neither cut nor touch axes, the distance from center to each axis must exceed the radius.
<p><strong>Step 1:</strong> For (2,5) to be interior to circle x² + y² - 8x - 12y + p = 0, substitute the point:<br>4 + 25 - 16 - 60 + p < 0<br>-47 + p < 0<br>p < 47</p><p><strong>Step 2:</strong> Rewrite circle in standard form: (x-4)² + (y-6)² = 52 - p<br>Center: (4, 6), Radius: r = √(52 - p)</p><p><strong>Step 3:</strong> For circle not to cut/touch x-axis: distance from center to x-axis > radius<br>6 > √(52 - p)<br>36 > 52 - p<br>p > -16</p><p><strong>Step 4:</strong> For circle not to cut/touch y-axis: distance from center to y-axis > radius<br>4 > √(52 - p)<br>16 > 52 - p<br>p > -36</p><p><strong>Step 5:</strong> Combining all conditions: p < 47 AND p > -16 AND p > -36<br>The binding constraint is: <strong>-16 < p < 47</strong></p><p>∴ Answer: A</p>
Correct Answer: A