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Question Number 83235 by peter frank last updated on 28/Feb/20

Commented by peter frank last updated on 28/Feb/20

$$chelp$$

Commented by john santu last updated on 29/Feb/20

$$\left(\mathrm{c}\right)\mathrm{let}\mathrm{the}\mathrm{circle}\mathrm{equation}\mathrm{is}$$$${(\mathrm{x}-\mathrm{p})}^2+{(\mathrm{y}-\mathrm{q})}^2={\mathrm{r}}^2$$$${(\mathrm{y}-\mathrm{q})}^2={\mathrm{r}}^2-{\mathrm{x}}^2+2\mathrm{px}-{\mathrm{p}}^2(\ast )$$$$\mathrm{let}\mathrm{vector}\mathrm{AB}=(\mathrm{x}-\mathrm{p}+\mathrm{r},\mathrm{y}-\mathrm{q})$$$$\mathrm{vector}\mathrm{BC}=(\mathrm{p}+\mathrm{r}-\mathrm{x},\mathrm{q}-\mathrm{y})$$$$\mathrm{and}\measuredangle \mathrm{ABC}=\theta$$$$\mathrm{BA}\bullet \mathrm{BC}=(\mathrm{p}-\mathrm{r}-\mathrm{x}).(\mathrm{p}+\mathrm{r}-\mathrm{x})+{(\mathrm{q}-\mathrm{y})}^2$$$$={\mathrm{p}}^2-{\mathrm{r}}^2-\mathrm{px}+\mathrm{rx}-\mathrm{px}-\mathrm{rx}+{\mathrm{x}}^2+$$$${\mathrm{r}}^2-{\mathrm{x}}^2+2\mathrm{px}-{\mathrm{p}}^2=0$$$$\mathrm{since}\mathrm{BA}\bullet \mathrm{BC}=0,\mathrm{then}\cos \theta =\cos \frac{\pi }{2}$$$$\mathrm{therefore}\mathrm{we}\mathrm{get}\theta =\frac{\pi }{2}$$

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