Question Number 178833 by Spillover last updated on 22/Oct/22
Answered by cortano1 last updated on 22/Oct/22
$$\left(\mathrm{b}\right)\:\mathrm{L}_{\mathrm{3}} \equiv\:\mathrm{L}_{\mathrm{2}} +\mathrm{k}\left(\mathrm{L}_{\mathrm{1}} −\mathrm{L}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{10x}−\mathrm{12y}+\mathrm{40}+\mathrm{k}\left(\mathrm{8x}+\mathrm{8y}−\mathrm{44}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{4k}−\mathrm{5}\right)\mathrm{x}+\mathrm{2}\left(\mathrm{4k}−\mathrm{6}\right)\mathrm{y}+\mathrm{40}−\mathrm{44k}=\mathrm{0} \\ $$$$\mathrm{whose}\:\mathrm{radius}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:\sqrt{\left(\mathrm{4k}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{4k}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{44k}−\mathrm{40}}=\mathrm{4} \\ $$$$\mathrm{let}\:\mathrm{4k}=\lambda \\ $$$$\Rightarrow\left(\lambda−\mathrm{5}\right)^{\mathrm{2}} +\left(\lambda−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{11}\lambda−\mathrm{40}=\mathrm{16} \\ $$$$\Rightarrow\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{11}\lambda+\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\lambda_{\mathrm{1}} =\mathrm{5}=\mathrm{4k}\:\:\&\:\lambda_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4k} \\ $$$$\Rightarrow\mathrm{L}_{\mathrm{3}} \equiv\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2y}−\mathrm{15}=\mathrm{0}\:\mathrm{or} \\ $$$$\Rightarrow\mathrm{L}_{\mathrm{3}} \equiv\:\mathrm{2x}^{\mathrm{2}} +\mathrm{2y}^{\mathrm{2}} −\mathrm{18x}−\mathrm{22y}+\mathrm{69}\:=\mathrm{0} \\ $$
Commented by cortano1 last updated on 22/Oct/22
Answered by Spillover last updated on 26/Dec/22
Answered by Spillover last updated on 26/Dec/22
Answered by Spillover last updated on 26/Dec/22
Answered by Spillover last updated on 26/Dec/22
Answered by Spillover last updated on 26/Dec/22
