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

From the given data (-6, 5) is the Orthocentre of triangle ABC

$$\begin{gathered} \mathrm{Given} \mathrm{vertices} \mathrm{A}=(3, 2) \mathrm{B}=(2, 1) \\ Let C=(x,y) \\ slope of BC\left(m_1\right)=\frac{y-1}{x-2} \\ slope of AH\left(m_2\right)=\frac{-1}{3} \\ {m}_1.m_2=-1\Rightarrow \frac{y-1}{x-2}\left(\frac{-1}{3}\right)=-1 \\ \Rightarrow 3x-y-5=0\to \left(1\right) \\ slope of AC\left(m_3\right)=\frac{y-2}{x-3} \\ slope of BH\left(m_4\right)=\frac{-1}{2} \\ {m}_3.m_4=-1 \\ \Rightarrow 2x-y-4=0\to \left(2\right) \\ By solving \left(1\right) and \left(2\right) we get C=(1,-2) \end{gathered}$$