thora sa maths pd lo ab xD________________________

Questions

August 2021 1 30 Report

thora sa maths pd lo ab xD

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centre of the circle passing through points (7,-5),(3,-7),(3,3) is ??
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Answers & Comments

BRAINLYADDICTOR

★FIND:

$\bold{Centre\:of\:the\:circle\:O(x, y)}$

★GIVEN,

Three points are

A(7,-5),B(3,-7),C(3,3)

★SOLUTION:

$distance \: b/w \:OA\\ = \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } \\ = \sqrt{(7 - x) {}^{2} + ( - 5 - y) {}^{2} } \\ similarly. \\ OB = \sqrt{(3 - x) {}^{2} + ( - 7 - y) {}^{2} }$

$OC = \sqrt{(3 - x) {}^{2} + (3 - y) {}^{2} }$

Radius of the circle are equal

SO,

OA=OB & OB=OC

$\sqrt{(7 - x) {}^{2} + ( - 5 - y) {}^{2} } =\\ \sqrt{(3 - x) {}^{2} + ( - 7 - y) {}^{2} } \\ b.s \: square \: root \: cancelling \\ =>(7 - x) {}^{2} + ( - 5 - y) {}^{2} = (3 - x) {}^{2} + ( - 7 - y) {}^{2} \\ =>49 + x {}^{2} - 14x + 25 + y {}^{2} + 10y =9 + x {}^{2} - 6x + 49 + y {}^{2} + 14y \\=> x {}^{2} + y {}^{2} - 14x + 10y + 74 = x {}^{2} + y {}^{2} - 6x + 14y + 58 \\ => - 14x + 10y + 74 = - 6x + 14y +58 \\ = > - 6x + 14x + 14y - 10y + 58 - 74 = 0 \\ = > 8x + 4y - 16 = 0 \\ = > 2x + y - 4 = 0...eq1\\ similarly. \\ OB=OC$

$\sqrt{(3 - x) {}^{2} + ( - 7 - y) {}^{2} } = \sqrt{(3 - x) {}^{2} + (3 - y) {}^{2} } \\ =>(3 - x) {}^{2} + ( - 7 - y) {}^{2} = (3 - x) {}^{2} + (3 - y) {}^{2} \\=> 9 + x {}^{2} - 6x + 49 + y {}^{2} + 14y = 9 + x {}^{2} - 6x + 9 + y {}^{2} - 6y \\=> - 6x + 14y + 58 = - 6x - 6y + 18 \\=> 14y + 58 = - 6y + 18 \\ = > 14y + 6y = 18 - 58 \\ = > 20y = - 40 \\ = > y = - 40/20 \\ = > y = - 2$

$sub. \: y = - 2 \: in \: eq1 \\ = > 2x + ( - 2) - 4 = 0 \\ = > 2x - 2 - 4 = 0 \\ = > 2x - 6 = 0 \\ = > 2x = 6 \\ = > x = 6/2 \\ = > x = 3 \\ so,(x,y) = (3, - 2)$