Circles
Position of a point with respect to a circle
Grade 11

Question:

<p>The point \((1, 4)\) is inside the circle \(S\) whose equation is of the form \(x^2 + y^2 - 6x - 10y + k = 0\), \(k\) being an arbitrary constant. Find the possible values of \(k\) if the circle \(S\) neither touches the axes nor cuts them.</p>

Step-by-Step Solution

Key Concept: For a circle to neither touch nor cut the axes, the distance from center to each axis must be strictly greater than the radius. The point (1,4) being inside means the circle's equation evaluated at that point must be negative.
<p><strong>Step 1:</strong> Rewrite circle in standard form.</p><p>x² + y² - 6x - 10y + k = 0</p><p>(x - 3)² + (y - 5)² - 9 - 25 + k = 0</p><p>(x - 3)² + (y - 5)² = 34 - k</p><p>Center: C(3, 5), Radius: r = √(34 - k)</p><p><strong>Step 2:</strong> Apply condition: Point (1, 4) is inside the circle.</p><p>Substituting (1, 4) in x² + y² - 6x - 10y + k:</p><p>1 + 16 - 6 - 40 + k < 0</p><p>k < 29</p><p><strong>Step 3:</strong> Circle must not touch or cut x-axis.</p><p>Distance from C(3, 5) to x-axis = 5</p><p>For no intersection: 5 > √(34 - k)</p><p>25 > 34 - k</p><p>k > -9</p><p><strong>Step 4:</strong> Circle must not touch or cut y-axis.</p><p>Distance from C(3, 5) to y-axis = 3</p><p>For no intersection: 3 > √(34 - k)</p><p>9 > 34 - k</p><p>k > 25</p><p><strong>Step 5:</strong> Combine all conditions.</p><p>From (1,4) inside: k < 29</p><p>From y-axis condition: k > 25</p><p>From x-axis condition: k > -9 (already satisfied by k > 25)</p><p>∴ Answer: 25 < k < 29, i.e., k ∈ (25, 29)</p>
Correct Answer: 25 < k < 29, i.e., k ∈ (25, 29)

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