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Question Number 134287 by EDWIN88 last updated on 02/Mar/21

$$\mathrm{The}\mathrm{straight}\mathrm{line}\mathrm{x}+\mathrm{y}-1=0\mathrm{meets}\mathrm{the}\mathrm{cicle}$$$${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-8\mathrm{y}=0\mathrm{at}\mathrm{A}\mathrm{and}\mathrm{B}.\mathrm{Then}\mathrm{find}$$$$\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{of}\mathrm{which}\mathrm{AB}\mathrm{is}\mathrm{diameter}$$

Answered by bramlexs22 last updated on 02/Mar/21

$$\mathrm{let}\mathrm{the}\mathrm{requires}\mathrm{equation}\mathrm{of}$$$$\mathrm{circle}\mathrm{is}{\mathrm{C}}_1+\lambda \ell =0;\mathrm{then}$$$$\Rightarrow x^2+y^2-6x-8y+\lambda (x+y-1)=0$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+(\lambda -6)\mathrm{x}+(\lambda -8)\mathrm{y}-\lambda =0$$$$\mathrm{with}\mathrm{center}\mathrm{point}\mathrm{at}(\frac{6-\lambda }{2},\frac{8-\lambda }{2})$$$$\mathrm{since}\mathrm{the}\mathrm{center}\mathrm{point}\mathrm{lie}\mathrm{on}\mathrm{line}$$$$\mathrm{x}+\mathrm{y}-1=0\Rightarrow \mathrm{we}\mathrm{find}\frac{6-\lambda }{2}+\frac{8-\lambda }{2}=1$$$$\Rightarrow 14-2\lambda =2;\lambda =6.$$$$\mathrm{therefore}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}$$$$\mathrm{of}\mathrm{which}\mathrm{AB}\mathrm{is}\mathrm{diameter}\equiv$$$${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{y}-6=0$$$$$$

Commented by bramlexs22 last updated on 02/Mar/21

Answered by EDWIN88 last updated on 02/Mar/21

$$\left(1\right)\mathrm{x}=1-\mathrm{y}\Rightarrow {\mathrm{x}}^2=1-2\mathrm{y}+{\mathrm{y}}^2$$$$\left(2\right)1-2\mathrm{y}+{\mathrm{y}}^2+{\mathrm{y}}^2-6(1-\mathrm{y})-8\mathrm{y}=0$$$${2\mathrm{y}}^2-4\mathrm{y}-5=0$$$${\mathrm{y}}^2-2\mathrm{y}-\frac{5}{2}=0$$$${(\mathrm{y}-1)}^2=\frac{7}{2};\{\begin{array}{l}{\mathrm{y}}_1=1+\frac{\sqrt{7}}{\sqrt{2}}\to {\mathrm{x}}_1=-\frac{\sqrt{7}}{\sqrt{2}}\\ {\mathrm{y}}_2=1-\frac{\sqrt{7}}{\sqrt{2}}\to {\mathrm{x}}_2=\frac{\sqrt{7}}{\sqrt{2}}\end{array}$$$$\mathrm{therefore}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{which}\mathrm{AB}\mathrm{is}$$$$\mathrm{diameter}\Rightarrow (\mathrm{x}-\frac{\sqrt{7}}{\sqrt{2}})(\mathrm{x}+\frac{\sqrt{7}}{\sqrt{2}})+(\mathrm{y}-\left(\frac{\sqrt{2}+\sqrt{7}}{\sqrt{2}}\right))(\mathrm{y}-\left(\frac{\sqrt{2}-\sqrt{7}}{\sqrt{2}}\right))=0$$$$\iff {\mathrm{x}}^2-\frac{7}{2}+{\mathrm{y}}^2-\left(\frac{\sqrt{2}-\sqrt{7}}{\sqrt{2}}\right)\mathrm{y}-\left(\frac{\sqrt{2}+\sqrt{7}}{\sqrt{2}}\right)\mathrm{y}+\left(\frac{2-7}{2}\right)=0$$$$\iff {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{y}-6=0$$$$$$

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