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Find radius of the circle \( 2 x^{2}+2 y^{2}+p x+p y \). |
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Answer» The given equation of circle is 2x2 + 2y2 + px + py = 0 ⇒ x2 + y2 + p/2 x + p/2 y = 0 (Dividing both sides by 2) ⇒ (x2 + p/2 x + p2/16) + (y2 + p/2 y + p2/16) - p2/16 - p2/16 = 0 (Adding and subtracting p2/16 in L.H.S) ⇒ (x + p/4)2 + (y + p/4)2 = 2p2/16 = p2/8 By comparing (x - h)2 + (y - k)2 = r2, we get h = \(\frac{-p}4\) and k = \(\frac{-p}4\) and r2 = p2/8 ⇒ r = r/2√2 Hence, radius of the circle is r = p/2√2 |
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