Question

दिखाइए कि शीर्षो \( (2,-1),(3,2) \) तथा \( (3,8) \) वाल त्रिभुज पूर्णतया वृत्त \( x^{2}+y^{2}-10 x-6 y-3=0 \) के भीतर है।

Answer 1

Let P(x, y) = x² + y² - 10x - 6y - 3

P(2, -1) = 4 + 1 - 20 + 6 - 3 = 11 - 23 = -12 < 0

P(3, 2) = 9 + 4 - 30 - 12 - 3 = 13 - 45 = -32 < 0

P(3, 8) = 9 + 64 - 30 - 48 - 3 = 73 - 81 = -8 < 0

Hence, point (2, -1), (3, 2) & (3, 8) will lie inside the circle. Therefore triangle which made by vertices (2, -1), (3, 2) & (3, 8) will completely lie inside the given circle.