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If $\mathrm{P}(6,1)$ be the orthocentre of the triangle whose vertices are $\mathrm{A}(5,-2), \mathrm{B}(8,3)$ and $\mathrm{C}(\mathrm{h}, \mathrm{k})$, then the point $C$ lies on the circle:
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The correct answer is:
$x^2+y^2-65=0$
Slope of $A D=3$
Slope of $\mathrm{BC}=-\frac{1}{3}$
equation of $\mathrm{BC}=3 \mathrm{y}+\mathrm{x}-17=0$
slope of $\mathrm{BE}=1$
Slope of $\mathrm{AC}=-1$
equation of $\mathrm{AC}$ is $\mathrm{x}+\mathrm{y}-3=0$
point $C$ is $(-4,7)$
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