Question:
The radius of the circle passing through the foci of the ellipse \( \frac{x^2}{4} + \frac{4y^2}{7} = 1 \) and having its centre at \( \left(\frac{1}{2}, 2\right) \) is
The radius of the circle passing through the foci of the ellipse \( \frac{x^2}{4} + \frac{4y^2}{7} = 1 \) and having its centre at \( \left(\frac{1}{2}, 2\right) \) is
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Always reduce ellipse into standard form before identifying $a, b, c$.
Updated On: Apr 23, 2026
- $\sqrt{5}$
- $3$
- $\sqrt{12}$
- $\frac{7}{2}$
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The Correct Option is A
Solution and Explanation
Concept:
Ellipse standard form gives $c^2 = a^2 - b^2$ and foci at $(\pm c, 0)$.
Step 1: Convert to standard form.
\[ \frac{x^2}{4} + \frac{y^2}{\frac{7}{4}} = 1 \Rightarrow a^2 = 4,\ b^2 = \frac{7}{4} \]
Step 2: Find focal distance.
\[ c^2 = 4 - \frac{7}{4} = \frac{9}{4} \Rightarrow c = \frac{3}{2} \] Foci: $\left(\pm \frac{3}{2}, 0\right)$
Step 3: Use distance formula.
\[ r = \sqrt{\left(\frac{3}{2} - \frac{1}{2}\right)^2 + (0-2)^2} = \sqrt{1 + 4} = \sqrt{5} \] Conclusion:
Radius = $\sqrt{5}$
Step 1: Convert to standard form.
\[ \frac{x^2}{4} + \frac{y^2}{\frac{7}{4}} = 1 \Rightarrow a^2 = 4,\ b^2 = \frac{7}{4} \]
Step 2: Find focal distance.
\[ c^2 = 4 - \frac{7}{4} = \frac{9}{4} \Rightarrow c = \frac{3}{2} \] Foci: $\left(\pm \frac{3}{2}, 0\right)$
Step 3: Use distance formula.
\[ r = \sqrt{\left(\frac{3}{2} - \frac{1}{2}\right)^2 + (0-2)^2} = \sqrt{1 + 4} = \sqrt{5} \] Conclusion:
Radius = $\sqrt{5}$
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