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Home » Using the formula for the area in terms of the inradius, we have: - IQnection

Using the formula for the area in terms of the inradius, we have: - IQnection

📅 May 5, 2026 👤 scraface

May 05, 2026

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64x^2 + 6400 - 100x^2 = 6400
-36x^2 = 0 \Rightarrow x^2 = 0
Wait — this suggests only \( x = 0 \), but let's recheck:
Correct substitution:
\frac{x^2}{100} + 1 - \frac{x^2}{64} = 1
x^2\left( \frac{1}{100} - \frac{1}{64} \right) = 0
x^2 \left( \frac{64 - 100}{6400} \right) = 0 \Rightarrow x^2(-36/6400)...
So \( x = 0 \) is the only solution — but this implies the circle is t...
At \( x = 0 \): \( y^2 = 64 \), so \( y = \pm8 \).
Check in ellipse: \( 0 + \frac{64}{64} = 1 \) — yes.

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