Ellipse
Circle and Ellipse
Grade 11
Question:
<p>The equation of the circle passing through the foci of the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\), and having centre at \((0, 3)\) is:</p>
<p>(a) \(x^2 + y^2 - 6y + 5 = 0\)</p>
<p>(b) \(x^2 + y^2 - 6y - 7 = 0\)</p>
<p>(c) \(x^2 + y^2 - 6y + 7 = 0\)</p>
<p>(d) \(x^2 + y^2 - 6y - 5 = 0\)</p>
Step-by-Step Solution
Key Concept: Find the foci of the ellipse using \(c^2 = a^2 - b^2\), then use the distance formula from centre to focus to find the radius.
<p>The correct answer is (b) \(x^2 + y^2 - 6y - 7 = 0\)</p><p><strong>Step 1:</strong> For the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\), we have \(a^2 = 16\), \(b^2 = 9\), so \(c^2 = a^2 - b^2 = 7\) and \(c = \sqrt{7}\).</p><p><strong>Step 2:</strong> The foci are at \((\pm\sqrt{7}, 0)\).</p><p><strong>Step 3:</strong> The circle has centre \((0, 3)\) and passes through \((\sqrt{7}, 0)\). Radius \(r = \sqrt{7 + 9} = 4\).</p><p><strong>Step 4:</strong> Circle equation: \(x^2 + (y-3)^2 = 16\) which simplifies to \(x^2 + y^2 - 6y - 7 = 0\).</p>
Correct Answer: b