Question

Find the circumcentre of the triangle with the vertices A (-3,4) , B (3,4) and C (-4,-3). What is the circumradius of the circle?

Answer 1

We want the equation of circle to pass through points $A(-3,4),B(3,4)$ and $(-4,3)$

The equation of a circle is $x^2+y^2+ax+by+c=0$

From the coordinates of the three given points we get three equations.

$\Rightarrow {(-3)}^2+4^2-3a+4b+c=0$

$\Rightarrow 3^2+4^2-3a+4b+c=0$

$\Rightarrow {(-4)}^2+3^2-4a+3b+c=0$

we can simplify the above equation -

$\Rightarrow \left[\begin{array}{ccc}-3& 4& 1\\ 3& 4& 1\\ -4& 3& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-25\\ -25\\ -25\end{array}\right]$

solving the above matrix equation using craner's rule $a=0,$ $b=0$ and $c=-25$

Putting there values $a,b$ and $c$ in general form we write required equation of circle $x^2+y^2-25=0$

On center radius form ${(x+0)}^2+{(y+0)}^2=25$

Center of circle is $(0,0)$ and radius $=5.$

Hence, the answer is $5.$