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Question Number 57174 by Tinkutara last updated on 31/Mar/19

Answered by ajfour last updated on 31/Mar/19

eq. of L y=3+rsin θ , x=2+rcos θ Q≡(2+Rcos θ,3+Rsin θ)≡(h,k) 3+r_A sin θ=12+r_A cos θ ⇒ r_A =(9/(sin θ−cos θ)) r_B =((19)/(sin θ−cos θ)) As r_A r_B =R^2 ⇒ ((171)/((sin θ−cos θ)^2 ))=R^2 or (k−3−h+2)^2 =171 ⇒ locus is (y−x−1)^2 =171 .

$$\mathrm{eq}.\:\mathrm{of}\:\mathrm{L} \\ $$$$\mathrm{y}=\mathrm{3}+\mathrm{rsin}\:\theta\:\:,\:\:\mathrm{x}=\mathrm{2}+\mathrm{rcos}\:\theta \\ $$$$\mathrm{Q}\equiv\left(\mathrm{2}+\mathrm{Rcos}\:\theta,\mathrm{3}+\mathrm{Rsin}\:\theta\right)\equiv\left(\mathrm{h},\mathrm{k}\right) \\ $$$$\mathrm{3}+\mathrm{r}_{\mathrm{A}} \mathrm{sin}\:\theta=\mathrm{12}+\mathrm{r}_{\mathrm{A}} \mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\mathrm{r}_{\mathrm{A}} =\frac{\mathrm{9}}{\mathrm{sin}\:\theta−\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:\:\:\:\mathrm{r}_{\mathrm{B}} =\frac{\mathrm{19}}{\mathrm{sin}\:\theta−\mathrm{cos}\:\theta} \\ $$$$\:\:\:\mathrm{As}\:\:\:\:\mathrm{r}_{\mathrm{A}} \mathrm{r}_{\mathrm{B}} =\mathrm{R}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{171}}{\left(\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right)^{\mathrm{2}} }=\mathrm{R}^{\mathrm{2}} \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\:\left(\mathrm{k}−\mathrm{3}−\mathrm{h}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{171} \\ $$$$\Rightarrow\:\:\mathrm{locus}\:\mathrm{is}\:\:\:\:\left(\mathrm{y}−\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{171}\:\:\:. \\ $$

Commented by Tinkutara last updated on 31/Mar/19

Thank you so much Sir!

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