ग्राफ \(y=\sqrt{16-x^2}\) किस आकृति का ऊपरी भाग है?
The graph \(y=\sqrt{16-x^2}\) is the upper part of which figure?
Explanation opens after your attempt
A. केंद्र ((0,0)) और त्रिज्या (4) वाला वृत्तcircle with centre ((0,0)) and radius (4)
Concept
Squaring gives \(x^2+y^2=16\) with \(y\ge0\). In exams, identify \(\sqrt{r^2-x^2}\) as an upper semicircle.
Why this answer is correct
The correct answer is A. केंद्र ((0,0)) और त्रिज्या (4) वाला वृत्त / circle with centre ((0,0)) and radius (4). Squaring gives \(x^2+y^2=16\) with \(y\ge0\). In exams, identify \(\sqrt{r^2-x^2}\) as an upper semicircle.
Exam Tip
वर्ग करने पर \(x^2+y^2=16\) और \(y\ge0\) मिलता है। परीक्षा में \(\sqrt{r^2-x^2}\) को ऊपरी अर्धवृत्त पहचानें।
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