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Let the orthocentre and centroid of a triangle be $A(-5,3)$ and $B(4,4)$ , respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is

Option: 1
$\sqrt{24620} / 358$

Option: 2
$\sqrt{12310} / 179$

Option: 3
$\sqrt{12410} / 358 \mid$

Option: 4
$\sqrt{6205} / 179$

Answers (1)

best_answer

The midpoint of AB is $(-5+4) / 2,(3+4) / 2)=(-0.5,3.5)$

The slope of the line passing through $(-0.5,3.5)$ and $(4,4)$

$(4-3.5) /(4-(-0.5))=0.1$

So, the equation of the line passing through A and B is

The slope of the line passing through A and B is $y-3=0.1(x+5)$

The slope of the line passing through A and B is $(-4-3) /(4+5)=-7 / 9$

So, the equation of the perpendicular bisector of AB passing through its midpoint is $y-3.5=(-9 / 7)(x+0.5)$

Now, the equation of the perpendicular bisector of BC passing through its midpoint, which is the midpoint of BC.

Since the centroid divides the median in the ratio $2:1$ , the coordinates of point C are:

$(24 / 3+1(-5 / 3), 24 / 3+13 / 3)=(7 / 3,10 / 3)$

The midpoint of BC is $((4+7 / 3) / 2,(4+10 / 3) / 2)=(19 / 6,22 / 6)=(19 / 6,11 / 3)$

The slope of the line passing through B and C is $(10 / 3-4) /(7 / 3-4)=-2 / 5$

So, the equation of the perpendicular bisector of BC passing through its midpoint is:

$y-11 / 3=(5 / 2)(x-19 / 6)$

The system of equations given by the two perpendicular bisectors:

$\begin{aligned} & \Rightarrow y=(-9 / 7) x+11 \\ \Rightarrow & y=(5 / 2) x-1 / 6 \\ \Rightarrow & x=123 / 179 \\ \Rightarrow & y=139 / 179 \end{aligned}$

So, the coordinates of the circumcentre C are $(123/179,139/179)$ $\begin{aligned} & \left.r=\sqrt{(}(123 / 179-(-5))^2+(139 / 179-3)^2\right) / 2 \\ & r=\sqrt{(12410) / 358 .} \\ & \end{aligned}$

The radius of the circle with AC as diameter is half the distance between A and C:

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