Circles
Centre of a circle
Grade 11
Question:
<p>The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle \(x^2 + y^2 = 9\) is</p>
<p>\(\left(\dfrac{3}{2}, \dfrac{1}{2}\right)\)</p>
<p>\(\left(\dfrac{1}{2}, \dfrac{3}{2}\right)\)</p>
<p>\(\left(\dfrac{1}{2}, \dfrac{1}{2}\right)\)</p>
<p>\(\left(\dfrac{1}{2}, -\sqrt{2}\right)\)</p>
Step-by-Step Solution
Key Concept: A circle passing through (0,0) and (1,0) has its center on the perpendicular bisector of the chord joining these points (the line x = 1/2). Use the tangency condition with x² + y² = 9 to find the y-coordinate by equating the distance between centers to the difference of radii.
<p><strong>Step 1:</strong> Since the circle passes through (0,0) and (1,0), its center lies on the perpendicular bisector of this chord, which is the line <strong>x = 1/2</strong>.</p><p><strong>Step 2:</strong> Let the center be C = (1/2, h). The radius is: r = √[(1/2)² + h²] = √(1/4 + h²)</p><p><strong>Step 3:</strong> The given circle x² + y² = 9 has center O = (0,0) and radius R = 3. Distance between centers: d = √(1/4 + h²)</p><p><strong>Step 4:</strong> For tangency, either d = r + R (external) or d = |R - r| (internal).</p><p><strong>Step 5 (Internal Tangency):</strong> √(1/4 + h²) = 3 - √(1/4 + h²)<br>2√(1/4 + h²) = 3<br>√(1/4 + h²) = 3/2<br>1/4 + h² = 9/4<br>h² = 2<br>h = ±√2</p><p><strong>Step 6 (External Tangency):</strong> √(1/4 + h²) = 3 + √(1/4 + h²) gives no solution (contradiction).</p><p>∴ Answer: <strong>D</strong> (1/2, √2) or (1/2, -√2)</p>
Correct Answer: D