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Question Number 59287 by pete last updated on 07/May/19

$$\mathrm{A}\mathrm{circle}{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-5=0\mathrm{with}\mathrm{centr}$$$$0\mathrm{is}\mathrm{cut}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{y}=2\mathrm{x}+5\mathrm{at}\mathrm{points}\mathrm{P}\mathrm{and}\mathrm{Q}.$$$$\mathrm{Show}\mathrm{that}\mathrm{QO}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{PO}.$$

Answered by tanmay last updated on 07/May/19

$$solvey=2x+5and{x}^2+y^2-2x-4y-5=0$$$$x^2+4x^2+20x+25-2x-4(2x+5)-5=0$$$$5x^2+10x=0$$$$x(x+2)=0$$$$whenx=0y=5$$$$whenx=-2y=1$$$$P(0,5)Q(-2,1)$$$$x^2-2x+1+y^2-4y+4-10=0$$$${(x-1)}^2+{(y-2)}^2=10$$$$centreO(1,2)$$$$m_1=slopeOP=\frac{5-2}{0-1}=-3$$$$m_2=slopeOQ=\frac{2-1}{1-(-2)}=\frac{1}{3}$$$$m_1\times m_2=-3\times \frac{1}{3}=-1$$$$soOP\mathrm{\perp }OQ$$

Commented by pete last updated on 07/May/19

$$\mathrm{thanks}\mathrm{very}\mathrm{much}\mathrm{sir}$$

Commented by tanmay last updated on 07/May/19

$$mostwelcome...$$

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