Question

(d) Find the centre of the circle passing through the points (3,0), (2, √5) and (-2√2, -1).​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{The given circle passes through}$

$\mathsf{(3,0),(2,\sqrt{5})\;and\;(-2\sqrt{2},-1)}$

$\underline{\textsf{To find:}}$

$\textsf{Centre of the circle passes through given 3 points}$

$\underline{\textsf{Solution:}}$

$\textsf{Let the given 3 points be}$

$\mathsf{A(3,0),B(2,\sqrt{5})\;and\;C(-2\sqrt{2},-1)}$

$\textsf{Let P(h,k) be the centre of the circle}$

$\mathsf{Then,\;PA=PB=PC}$

$\implies\mathsf{\sqrt{(h-3)^2+k^2}=\sqrt{(h-2)^2+(k-\sqrt{5})^2}=\sqrt{(h+2\sqrt{2})^2+(k+1)^2}}$

$\mathsf{Consider,}$

$\implies\mathsf{\sqrt{(h-3)^2+k^2}=\sqrt{(h-2)^2+(k-\sqrt{5})^2}}$

$\implies\mathsf{\sqrt{h^2+9-6h+k^2}=\sqrt{h^2+4-4h+k^2+5-2\sqrt{5}k}}$

$\implies\mathsf{h^2+9-6h+k^2=h^2+4-4h+k^2+5-2\sqrt{5}k}$

$\implies\mathsf{-6h=-4h-2\sqrt{5}k}$

$\implies\mathsf{-2h=-2\sqrt{5}k}$

$\implies\mathsf{h=\sqrt{5}k}$........(1)

$\mathsf{Also}$

$\implies\mathsf{\sqrt{(h-3)^2+k^2}=\sqrt{(h+2\sqrt{2})^2+(k+1)^2}}$

$\implies\mathsf{h^2+9-6h+k^2=h^2+8+4\sqrt{2}h+k^2+1+2k}$

$\implies\mathsf{-6h=4\sqrt{2}h+2k}$

$\implies\mathsf{(-6-4\sqrt{2})h=2k}$

$\implies\mathsf{(-6-4\sqrt{2})\sqrt{5}k=2k}$

$\implies\mathsf{((-6-4\sqrt{2})\sqrt{5}-2)k=0}$

$\implies\mathsf{k=0}$

$\mathsf{Put\;k=0\;in\;(1),\;h=0}$

$\therefore\boxed{\bf\textsf{Centre of the circle is (0,0)}}$

$\underline{\textsf{Find more:}}$

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