ABC is a triangle and the radical centre of t | vedclass

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Question

(NONE) The radical centre of the circles described on the sides of a triangle as diameters is the orthocentre of the triangle.
Let the orthocentre $H

Vedclass · Accepted Answer

$(9,22)$

Vedclass · Answer

(NONE) The radical centre of the circles described on the sides of a triangle as diameters is the orthocentre of the triangle.
Let the orthocentre $H = (-6,5)$.
Let $A = (3,2)$,$B = (2,1)$,and $C = (x,y)$.
The slope of $BC$ is $m_{BC} = \frac{y-1}{x-2}$.
Since $AH \perp BC$,the slope of $AH$ is $m_{AH} = \frac{5-2}{-6-3} = \frac{3}{-9} = -\frac{1}{3}$.
Therefore,$m_{BC} = -\frac{1}{m_{AH}} = 3$.
So,$\frac{y-1}{x-2} = 3 \implies y-1 = 3x-6 \implies 3x-y = 5$ $(1)$.
The slope of $AC$ is $m_{AC} = \frac{y-2}{x-3}$.
Since $BH \perp AC$,the slope of $BH$ is $m_{BH} = \frac{5-1}{-6-2} = \frac{4}{-8} = -\frac{1}{2}$.
Therefore,$m_{AC} = -\frac{1}{m_{BH}} = 2$.
So,$\frac{y-2}{x-3} = 2 \implies y-2 = 2x-6 \implies 2x-y = -4$ $(2)$.
Subtracting $(2)$ from $(1)$: $(3x-y) - (2x-y) = 5 - (-4) \implies x = 9$.
Substituting $x=9$ in $(1)$: $3(9) - y = 5 \implies 27 - y = 5 \implies y = 22$.
Thus,$C = (9,22)$.
Note: Given options do not contain the correct coordinate. Based on the calculation,the correct point is $(9,22)$.

$ABC$ is a triangle and the radical centre of the circles with $AB, BC, CA$ as the diameters is $(-6,5)$. If $A=(3,2)$ and $B=(2,1)$,then $C=$

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