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Question Number 31249 by Joel578 last updated on 04/Mar/18

$$2\mathrm{lines}\mathrm{through}\mathrm{the}\mathrm{point}A(5,1)\mathrm{are}\mathrm{tangent}$$$$\mathrm{to}\mathrm{the}\mathrm{circle}{x}^2+y^2-4x+6y+4=0$$$$\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{these}2\mathrm{lines}$$

Commented by Joel578 last updated on 04/Mar/18

$$\mathrm{Equation}\mathrm{of}\mathrm{line}\mathrm{through}(5,1)$$$$y-1=m(x-5)$$$$y=mx-5m+1$$$$$$$${(x-2)}^2+{(y+3)}^2=9$$$$\mathrm{Equation}\mathrm{of}\mathrm{tangent}\mathrm{lines}\mathrm{with}\mathrm{slope}m$$$$y+3=m(x-2)\pm 3\sqrt{1+m^2}$$$$\mathrm{Substitute}(5,1)$$$$4=3(m\pm \sqrt{1+m^2})$$$$\frac{4}{3}-m=\pm \sqrt{1+m^2}$$$$\frac{16}{9}-\frac{8m}{3}+m^2=1+m^2$$$$\frac{16}{9}-1=\frac{8m}{3}$$$$16-9=24m$$$$m=\frac{7}{24}$$$$$$$$\mathrm{Equation}\mathrm{of}\mathrm{tangent}\mathrm{line}$$$$y=\frac{7}{24}x-\frac{11}{24}$$

Commented by Joel578 last updated on 04/Mar/18

$$Ionlyfound1solution.{What}^{\prime }swrong$$$$withmyanswer?$$

Commented by ajfour last updated on 04/Mar/18

$$Alsom=\mathrm{\infty }$$$$soanothertangentis$$$$x-5=0.$$

Commented by Joel578 last updated on 05/Mar/18

$$\mathrm{How}\mathrm{do}\mathrm{I}\mathrm{know}\mathrm{that}\mathrm{another}m=\mathrm{\infty }$$$$\mathrm{especially}\mathrm{without}\mathrm{using}\mathrm{graphing}\mathrm{tool}?$$

Commented by ajfour last updated on 05/Mar/18

$$whenyoucanceltermof{m}^2,$$$$youshouldrealisethatm=\pm \mathrm{\infty }is$$$$ofcoursetheroots.Forexample$$$$m^2={(m-2)}^2$$$$\Rightarrow m^2=m^2-4m+4$$$$thiseq.hasroots$$$$m=\pm \mathrm{\infty }andm=1.$$

Commented by Joel578 last updated on 05/Mar/18

$$Sirwouldyouliketoexplainindetailedway?$$$$I^{\prime }mstill{don}^{\prime }tknowhowcanIsolveif$$$$Ichangedy\iff x$$

Commented by Joel578 last updated on 05/Mar/18

$$Okay.Thankyouverymuch$$

Commented by MJS last updated on 05/Mar/18

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{equation}y=kx+d\mathrm{for}$$$$\mathrm{any}\mathrm{vertical}\mathrm{line}$$$$\mathrm{remember}k=\tan \alpha ,\alpha =90°\Rightarrow \tan \alpha =\pm \mathrm{\infty }$$$$\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{example},\mathrm{if}\mathrm{you}\mathrm{change}$$$$x\iff y{\mathrm{you}}^{\prime }\mathrm{d}\mathrm{get}\mathrm{both}\mathrm{solutions}$$

Commented by MJS last updated on 05/Mar/18

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{enough}\mathrm{to}\mathrm{change}\mathrm{the}\mathrm{side}\mathrm{of}m$$$$m(y-1)=x-5$$$$x=my-m+5$$$$x-2=m(y+3)\pm 3\sqrt{1+m^2}$$$$3=4m\pm 3\sqrt{1+m^2}$$$$3-4m=\pm 3\sqrt{1+m^2}$$$$9-24m+16m^2=9+9m^2$$$$7m^2-24m=0$$$$m_1=0$$$$m_2=\frac{24}{7}$$$$t_1:x=0y-0+5\Rightarrow x=5$$$$t_2:x=\frac{24}{7}y-\frac{24}{7}+5\Rightarrow x=\frac{24}{7}y+\frac{11}{7}$$$$\Rightarrow y=\frac{7}{24}x-\frac{11}{24}$$$$\mathrm{your}\mathrm{method}\mathrm{cannot}\mathrm{show}\mathrm{a}$$$$\mathrm{vertical}\mathrm{tangent},\mathrm{mine}\mathrm{cannot}$$$$\mathrm{show}\mathrm{a}\mathrm{horizontal}\mathrm{one}$$$$\mathrm{so}\mathrm{if}\mathrm{one}\mathrm{tangent}\mathrm{is}\mathrm{vertical}$$$$\mathrm{and}\mathrm{the}\mathrm{other}{\mathrm{one}}^{\prime }\mathrm{s}\mathrm{horizontal},$$$$\mathrm{we}\mathrm{might}\mathrm{need}\mathrm{both}$$

Commented by Joel578 last updated on 06/Mar/18

$$Okay,thankyoufortheexplanation$$

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