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| 1. |
Find the centre of a circle passing through the points (6,-6) (3,-7) and(3,3) |
| Answer» Let the given points are\xa0A(6,-6), B(3,-7), C(3,3)And P(x,y)\xa0be the centre of the circle. So, AP = BP = CP (radii\xa0of the circle)Taking AP = BP and\xa0Squaring both sides, we get,{tex}\\Rightarrow A P ^ { 2 } = B P ^ { 2 }{/tex}{tex}\\Rightarrow ( x - 6 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = ( x - 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 }{/tex}(by using distance formula){tex}\\Rightarrow x ^ { 2 } - 12 x + 36 + y ^ { 2 } + 12 y + 36{/tex}\xa0={tex}x ^ { 2 } - 6 x + 9 + y ^ { 2 } + 14 y + 49{/tex}{tex}\\Rightarrow - 12 x + 6 x + 12 y - 14 y + 72 - 58 = 0{/tex}{tex}\\Rightarrow - 6 x - 2 y + 14 = 0{/tex}{tex}\\Rightarrow 3 x + y - 7 = 0{/tex}\xa0……….(i)Again, taking BP = CP and squaring both sides, we get,{tex}\\Rightarrow B P ^ { 2 } = C P ^ { 2 }{/tex}{tex}\\Rightarrow ( x - 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = ( x - 3 ) ^ { 2 } + ( y - 3 ) ^ { 2 }{/tex}{tex}\\Rightarrow x ^ { 2 } - 6 x + 9 + y ^ { 2 } + 14 y + 49{/tex}\xa0={tex}x ^ { 2 } - 6 x + 9 + y ^ { 2 } - 6 y + 9{/tex}{tex}\\Rightarrow - 6 x + 6 x + 14 y + 6 y + 58 - 18 = 0{/tex}{tex}\\Rightarrow 20 y + 40 = 0{/tex}{tex}\\Rightarrow y = - 2{/tex}Putting the value of y\xa0in eq. (i),{tex}3x + y - 7 = 0{/tex}{tex}\\Rightarrow 3 x = 9{/tex}{tex}\\Rightarrow x = 3{/tex}Hence the coordinates of P i.e. centre of circle are (3,-2). | |