In $\triangle A B C, \angle A=90^{\circ}$ and co-ordinates of points $B$ and $C$ are $(2,-4)$ and $(1,5)$. Then the equation of the circumcircle of $\triangle A B C$ is

Question

In $\triangle A B C, \angle A=90^{\circ}$ and co-ordinates of points $B$ and $C$ are $(2,-4)$ and $(1,5)$. Then the equation of the circumcircle of $\triangle A B C$ is, $x^2+y^2+3 x+y+18=0$, $x^2+y^2-3 x+y-18=0$, $x^2+y^2-3 x-y-18=0$, $x^2+y^2+3 x-y+18=0$

MARKS App · Accepted Answer

It is given that in $\triangle A B C, \angle A=90^{\circ}$, so equation of circumcircle of $\triangle A B C$, where $B(2,-4)$ and $C(1,5)$ $\because B$ and $C$ are end points of diameter of the circumcircle of $\triangle A B C$, so equation of circumcircle is $\begin{array}{rlrl} \Rightarrow & (x-2)(x-1)+(y+4)(y-5) =0 \ \Rightarrow & x^2+y^2-3 x-y-18 =0 \end{array}$ Hence, option (c) is correct.

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