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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

c help

$${c}\:{help} \\ $$

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

(c) let the circle equation is (x−p)^2 +(y−q)^2 = r^2 (y−q)^2 = r^2 −x^2 +2px−p^2 (∗) let vector AB = (x−p+r, y−q ) vector BC = (p+r−x, q−y) and ∡ABC = θ BA • BC = (p−r−x).(p+r−x)+(q−y)^2 = p^2 −r^2 −px+rx−px−rx+x^(2 ) + r^2 −x^2 +2px−p^2 = 0 since BA•BC = 0 , then cos θ = cos (π/2) therefore we get θ = (π/2)

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

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