JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 15)
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 $$\mathrm{C}$$ lies on the circle :
$$x^2+y^2-74=0$$
$$x^2+y^2-65=0$$
$$x^2+y^2-61=0$$
$$x^2+y^2-52=0$$
Explanation
_6th_April_Evening_Shift_en_15_1.png)
Slope of $$B F=\left(m_{B F}\right)=\frac{3-1}{8-6}=1$$
$$\begin{aligned} & m_{A C}=\frac{k+2}{h-5}=-1(A C \perp B F) \\ & \Rightarrow k+2=5-h \\ & \Rightarrow k=3-h \quad \ldots \text { (i) } \\ & m_{A D}=\frac{1+2}{6-5}=3 \\ & m_{B C}=\frac{k-3}{h-8}=-\frac{1}{3}(A D \perp B C) \\ & \Rightarrow 3 k-9=8-h \quad \ldots \text { (ii) } \end{aligned}$$
From (i) & (ii)
$$h=-4, k=7$$
Now, $$h^2+k^2=16+49=65$$
$$h^2+k^2-65=0$$
Locus is $$x^2+y^2-65=0$$
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