<p>If the point (1, 4) lies inside the circle \(x^2 + y^2 - 6x - 10y + p = 0\) and the circle does not touch or intersect the coordinate axes, then the set of all possible values of \(p\) is the interval:</p>
Step-by-Step Solution
Key Concept: A point lies inside a circle if substituting its coordinates into the circle equation gives a negative result (since the general form equals 0 on the circle). Additionally, a circle doesn't touch/intersect the axes when both the distance from center to x-axis and y-axis exceed the radius.
<p><strong>Step 1: Rewrite in standard form</strong></p><p>Circle: x² + y² - 6x - 10y + p = 0</p><p>Complete the square: (x-3)² + (y-5)² = 9 + 25 - p = 34 - p</p><p>Center C = (3, 5), radius r = √(34 - p)</p><p><strong>Step 2: Condition for point (1,4) inside the circle</strong></p><p>Substitute (1,4): 1 + 16 - 6 - 40 + p < 0</p><p>p - 29 < 0 ⟹ <strong>p < 29</strong></p><p><strong>Step 3: Circle doesn't intersect x-axis</strong></p><p>Distance from center to x-axis = 5</p><p>For no intersection: 5 > √(34 - p)</p><p>25 > 34 - p ⟹ <strong>p > 9</strong></p><p><strong>Step 4: Circle doesn't intersect y-axis</strong></p><p>Distance from center to y-axis = 3</p><p>For no intersection: 3 > √(34 - p)</p><p>9 > 34 - p ⟹ <strong>p > 25</strong></p><p><strong>Step 5: Combine all conditions</strong></p><p>p < 29 AND p > 9 AND p > 25</p><p>∴ Answer: <strong>(25, 29)</strong></p>
Correct Answer: D