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Question

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

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Solution

Given that the circle is passing through points (5, -8), (2, -9) and (2,1).

Let the center of the circle be (x,y) which will be equidistant from points (5, -8), (2, -9) and (2,1).

Taking last two points and applying distance formula,

square root of open parentheses x minus 2 close parentheses squared plus open parentheses y plus 9 close parentheses squared end root equals square root of open parentheses x minus 2 close parentheses squared plus open parentheses y minus 1 close parentheses squared end root s q u a r i n g space b o t h space s i d e s space open parentheses x minus 2 close parentheses squared plus open parentheses y plus 9 close parentheses squared equals open parentheses x minus 2 close parentheses squared plus open parentheses y minus 1 close parentheses squared open parentheses y plus 9 close parentheses squared equals open parentheses y minus 1 close parentheses squared y squared plus 18 y plus 81 equals y squared plus 1 minus 2 y 20 y equals negative 80 y equals negative 4

Taking first two points,

square root of open parentheses x minus 5 close parentheses squared plus open parentheses y plus 8 close parentheses squared end root to the power of blank equals square root of open parentheses x minus 2 close parentheses squared plus open parentheses y plus 9 close parentheses squared end root s q u a r i n g space b o t h space s i d e s space a n d space p u t t i n g space v a l u e space o f space y equals negative 4 open parentheses x minus 5 close parentheses squared plus open parentheses negative 4 plus 8 close parentheses squared equals open parentheses x minus 2 close parentheses squared plus open parentheses negative 4 plus 9 close parentheses squared x squared minus 10 x space plus 25 plus 16 equals x squared minus 4 x plus 4 plus 25 minus 6 x equals negative 12 x equals 2 space

hence the center of the circle (x,y)=(2,-4)


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