ग्राफ \(y=\sqrt{9-x^2}\) किस ज्यामितीय आकृति का ऊपरी भाग है?
The graph \(y=\sqrt{9-x^2}\) is the upper part of which geometric figure?
Explanation opens after your attempt
Correct Answer
A. केंद्र ((0,0)) और त्रिज्या (3) वाला वृत्तcircle with centre ((0,0)) and radius (3)
Concept
Squaring gives \(x^2+y^2=9\) with \(y\ge0\). In exams, identify \(\sqrt{r^2-x^2}\) as the upper semicircle.
Why this answer is correct
The correct answer is A. केंद्र ((0,0)) और त्रिज्या (3) वाला वृत्त / circle with centre ((0,0)) and radius (3). Squaring gives \(x^2+y^2=9\) with \(y\ge0\). In exams, identify \(\sqrt{r^2-x^2}\) as the upper semicircle.
Exam Tip
वर्ग करने पर \(x^2+y^2=9\) और \(y\ge0\) मिलता है। परीक्षा में \(\sqrt{r^2-x^2}\) को ऊपरी अर्धवृत्त पहचानें।
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