The centre and radius of the circumcircle of | vedclass

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(C) The equations of the sides of the triangle are $2x+3y=10$,$y=x$,and $y=0$ ($X$-axis).Solving these equations to find the vertices:$1$. Intersectio

Vedclass · Accepted Answer

$\left(\frac{5}{2}, \frac{-1}{2}\right), \sqrt{\frac{13}{2}}$

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(C) The equations of the sides of the triangle are $2x+3y=10$,$y=x$,and $y=0$ ($X$-axis).Solving these equations to find the vertices:$1$. Intersection of $y=x$ and $y=0$: $x=0, y=0$. So,$B(0,0)$.$2$. Intersection of $2x+3y=10$ and $y=0$: $2x=10 \Rightarrow x=5$. So,$C(5,0)$.$3$. Intersection of $2x+3y=10$ and $y=x$: $2x+3x=10$ $\Rightarrow 5x=10$ $\Rightarrow x=2, y=2$. So,$A(2,2)$.Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.Since it passes through $(0,0)$,$c=0$.Since it passes through $(5,0)$,$25+10g=0 \Rightarrow g=\frac{-25}{10}=\frac{-5}{2}$.Since it passes through $(2,2)$,$4+4+2(2)g+2(2)f=0 \Rightarrow 8+4g+4f=0$.Substituting $g=\frac{-5}{2}$: $8+4(\frac{-5}{2})+4f=0$ $\Rightarrow 8-10+4f=0$ $\Rightarrow 4f=2$ $\Rightarrow f=\frac{1}{2}$.The centre of the circle is $(-g, -f) = (\frac{5}{2}, \frac{-1}{2})$.The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(\frac{-5}{2})^2+(\frac{1}{2})^2-0} = \sqrt{\frac{25}{4}+\frac{1}{4}} = \sqrt{\frac{26}{4}} = \sqrt{\frac{13}{2}}$.

The centre and radius of the circumcircle of the triangle formed by the lines $2x+3y=10$,$y=x$ and the $X$-axis are respectively:

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