Circles
Circle through foci of ellipse
Grade 11
Question:
<p>The equation of the circle passing through the foci of the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1\), and having centre at (0, 3) is</p>
<p>\(x^2 + y^2 - 6y + 7 = 0\)</p>
<p>\(x^2 + y^2 - 6y - 5 = 0\)</p>
<p>\(x^2 + y^2 - 6y + 5 = 0\)</p>
<p>\(x^2 + y^2 - 6y - 7 = 0\)</p>
Step-by-Step Solution
Key Concept: Find the foci of the ellipse using c² = a² - b², then use the distance from center (0,3) to a focus to determine the radius of the circle.
<p><strong>Step 1:</strong> Identify ellipse parameters. From $\frac{x^2}{16} + \frac{y^2}{9} = 1$, we have $a^2 = 16$ and $b^2 = 9$, so $a = 4$ and $b = 3$.</p><p><strong>Step 2:</strong> Find the foci. Since $a^2 > b^2$, the major axis is along x-axis. Calculate $c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}$. The foci are at $(\pm\sqrt{7}, 0)$.</p><p><strong>Step 3:</strong> Find the radius. The circle has center $(0, 3)$ and passes through $(\sqrt{7}, 0)$ (or $(-\sqrt{7}, 0)$). The radius is: $r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.</p><p><strong>Step 4:</strong> Write the circle equation. With center $(0, 3)$ and radius $4$: $x^2 + (y - 3)^2 = 16$ or equivalently $x^2 + y^2 - 6y - 7 = 0$.</p><p>∴ Answer: D</p>
Correct Answer: D