Find the centre of a circle passing through(6,-6),(3,-7),(3,3)

1.	Find the centre of a circle passing through(6,-6),(3,-7),(3,3)
Answer» Let\xa0A → (6, –6), B\xa0→ (3, –7) and C\xa0→ (3, 3).Let the centre of the circle be I(x, y)Then, IA = IB = IC [By definition of a circle]{tex}\\Rightarrow{/tex} IA2 = IB2 = IC2{tex}\\Rightarrow{/tex} (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2Taking first two, we get(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2{tex}\\Rightarrow{/tex} x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49{tex}\\Rightarrow{/tex} 6x + 2y = 14{tex}\\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]Taking last two, we get(x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2{tex}\\Rightarrow{/tex} (y + 7)2 = (y - 3)2{tex}\\Rightarrow{/tex} (y + 7) = {tex}\\pm{/tex}(y-3)taking +e sign, we gety + 7 = y - 3{tex}\\Rightarrow{/tex} 7 = -3which is impossibleTaking -ve sign, we gety + 7 = -(y - 3){tex}\\Rightarrow{/tex} y + 7 = -y + 3{tex}\\Rightarrow{/tex} 2y = -4{tex}\\Rightarrow y = \\frac{{ - 4}}{2} = - 2{/tex}Putting y = -2 in equation (1), we get{tex}\\Rightarrow{/tex} 3x - 2 = 7{tex}\\Rightarrow{/tex} 3x = 9{tex}\\Rightarrow{/tex} x = 3Thus, I {tex}\\rightarrow{/tex} (3, -2)Hence, the centre of the circle is (3, -2).

Discussion