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22 Aug 2019, 01:19
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The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?
A. A
B. \(\sqrt{\frac{2A}{\pi}}\)
C. \(\sqrt{\frac{3A}{\pi}}\)
D. \(\frac{2A}{\pi}\)
E. \(\frac{3A}{\pi}\)
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22 Aug 2019, 01:19
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Bunuel
The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?
A. A
B. \(\sqrt{\frac{2A}{\pi}}\)
C. \(\sqrt{\frac{3A}{\pi}}\)
D. \(\frac{2A}{\pi}\)
E. \(\frac{3A}{\pi}\)
Similar question: https://gmatclub.com/forum/the-circle-w ... 44895.html A. A
B. \(\sqrt{\frac{2A}{\pi}}\)
C. \(\sqrt{\frac{3A}{\pi}}\)
D. \(\frac{2A}{\pi}\)
E. \(\frac{3A}{\pi}\)
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22 Aug 2019, 01:55
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Draw perpendiculars from C to the x and y axes. We now have a square with side \(r\)(radius of circle) and diagonal \(OC=\sqrt{2}r\)
We know that Area, \(A=\pi r^2\) and so \(r=\sqrt{\frac{A}{\pi}}\)
So \(OC=\sqrt{2}*\sqrt{\frac{A}{\pi}}=\sqrt{\frac{2A}{\pi}}\)
Answer is (B)
Posted from my mobile device
We know that Area, \(A=\pi r^2\) and so \(r=\sqrt{\frac{A}{\pi}}\)
So \(OC=\sqrt{2}*\sqrt{\frac{A}{\pi}}=\sqrt{\frac{2A}{\pi}}\)
Answer is (B)
Posted from my mobile device
