Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 79757 by mathocean1 last updated on 27/Jan/20

$$\mathrm{And}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{is}$$$$x^2+y^2-2x+4y=0.$$$$\left(T\right)\mathrm{is}\mathrm{his}\mathrm{his}\mathrm{tangent}\mathrm{line}\mathrm{at}\mathrm{M}(x_0;y_0)$$$$passingbyD(2;1).$$$$$$$$\mathrm{a})\mathrm{Show}\mathrm{that}\mathrm{y}\mathrm{verify}{\mathrm{y}}_0^2+{\mathrm{y}}_0=0$$$$\mathrm{b})\mathrm{deduct}\mathrm{others}{\mathrm{tangent}}^{\prime }\mathrm{s}$$$$\mathrm{equations}\mathrm{to}\mathrm{Circle}\mathrm{passing}\mathrm{by}\mathrm{D}.$$

Commented by john santu last updated on 27/Jan/20

$$\mathrm{center}\mathrm{the}\mathrm{circle}(1,-2),\mathrm{r}=\sqrt{5}$$$$\mathrm{let}\mathrm{tangent}\mathrm{line}\mathrm{y}=\mathrm{mx}+\mathrm{n}$$$$\mathrm{passing}\mathrm{D}(2,1)\Rightarrow 1=2\mathrm{m}+\mathrm{n}$$$$\mathrm{n}=1-2\mathrm{m}\Rightarrow \mathrm{y}=\mathrm{mx}+1-2\mathrm{m}$$$${\mathrm{d}}_{(1,-2)}=\mid \frac{\mathrm{m}+1+2+1-2\mathrm{m}}{\sqrt{{\mathrm{m}}^2+{(1-2\mathrm{m})}^2}}\mid$$$$\sqrt{5}=\mid \frac{4-\mathrm{m}}{\sqrt{{\mathrm{m}}^2+{(1-2\mathrm{m})}^2}}\mid$$$$5({\mathrm{m}}^2+{4\mathrm{m}}^2-4\mathrm{m}+1)={\mathrm{m}}^2-8\mathrm{m}+16$$$${25\mathrm{m}}^2-20\mathrm{m}+5-{\mathrm{m}}^2+8\mathrm{m}-16=0$$$${24\mathrm{m}}^2-12\mathrm{m}-11=0$$$$\mathrm{m}=\frac{12\pm \sqrt{144+1056}}{48}=\frac{12\pm 20\sqrt{3}}{48}$$$$\left(1\right)\mathrm{m}=\frac{3+5\sqrt{3}}{12}$$$$\mathrm{tangent}\mathrm{line}:\mathrm{y}=\left\{\frac{3+5\sqrt{3}}{12}\right\}\mathrm{x}+1-2\left\{\frac{3+\sqrt{5}}{12}\right\}$$$$\mathrm{y}=\left\{\frac{3+\sqrt{5}}{12}\right\}\mathrm{x}+\frac{6-2\sqrt{5}}{12}$$$$\mathrm{similarly}\mathrm{for}\mathrm{m}=\frac{3-\sqrt{5}}{12}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com