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Question Number 121355 by john santu last updated on 07/Nov/20

Answered by liberty last updated on 07/Nov/20

$$\mathrm{using}\mathrm{the}\mathrm{formula}\Rightarrow (\mathrm{x}-{\mathrm{x}}_1)(\mathrm{x}-{\mathrm{x}}_2)+(\mathrm{y}-{\mathrm{y}}_1)(\mathrm{y}-{\mathrm{y}}_2)=0$$$$\mathrm{where}\mathrm{the}\mathrm{point}({\mathrm{x}}_1,{\mathrm{y}}_1)\mathrm{\&}({\mathrm{x}}_2,{\mathrm{y}}_2)\mathrm{are}$$$$\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{diameter}\mathrm{circle}$$$$\iff (\mathrm{x}-\mathrm{a})(\mathrm{x}-10)+(\mathrm{y}-3)(\mathrm{y}-6)=0$$$$\mathrm{since}\mathrm{the}\mathrm{circle}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}$$$$(2,4)\mathrm{give}(2-\mathrm{a})(2-10)+(4-3)(4-6)=0$$$$\mathrm{we}\mathrm{get}-8(2-\mathrm{a})=2;\mathrm{a}=2+\frac{1}{4}=\frac{9}{4}$$$$\mathrm{so}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{is}$$$$(\mathrm{x}-\frac{9}{4})(\mathrm{x}-10)+(\mathrm{y}-3)(\mathrm{y}-6)=0$$$$\iff {\mathrm{x}}^2-\frac{49}{4}\mathrm{x}+\frac{90}{4}+{\mathrm{y}}^2-9\mathrm{y}+18=0$$$$\iff {(\mathrm{x}-\frac{49}{8})}^2+{(\mathrm{y}-\frac{9}{2})}^2=\frac{2401}{64}+\frac{81}{4}-\frac{90}{4}-18$$$$\iff {(\mathrm{x}-\frac{49}{8})}^2+{(\mathrm{y}-\frac{9}{2})}^2=\frac{2401-144-1152}{64}$$$$\iff {(\mathrm{x}-\frac{49}{8})}^2+{(\mathrm{y}-\frac{9}{2})}^2=\frac{1105}{64}$$$$\mathrm{and}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{diameter}\mathrm{passes}\mathrm{througt}$$$$\mathrm{the}\mathrm{points}\mathrm{A}(\frac{9}{4},3)\mathrm{and}\mathrm{B}(10,6)\mathrm{is}$$$${\Rightarrow }^{-\times \left(y\right)}{\left|\begin{array}{c}\frac{9}{4}3\\ 106\end{array}\right|}_{-\times \left(x\right)}\Rightarrow 3\mathrm{x}-\frac{31}{4}\mathrm{y}=30-\frac{186}{4}$$$$\Rightarrow 3\mathrm{x}-\frac{31}{4}\mathrm{y}+\frac{66}{4}=0\mathrm{or}12\mathrm{x}-31\mathrm{y}+66=0$$

Commented by john santu last updated on 07/Nov/20

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