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 August 30th, 2013, 04:58 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 The Equation of a Tangent to a Circle For this question, my final answer in (ii) almost matches that of the books answer. Can anyone confirm if I have the correct answer, or if the book is incorrect? Many thanks. Q. A circle has centre (3, 5) & touches the line $y=2x+4$. (i) Find the equation of this circle, (ii) Find the equation of the tangent to the circle at the point (1, 4). Attempt: (i) Distance between c & T: $\frac{|3(2)+5(-1)+4|}{\sqrt{2^2+(-1)^2}}=\frac{5}{\sqrt{5}}=\sqrt{5}$ Equation of circle: $(x-3)^2+(y-5)^2=\sqrt{5}\rightarrow (x-3)^2+(y-5)^2=5$ (ii) Equation of line at (1, 4): $y-4=m(x-5)\rightarrow mx-y+4-5m=0$ Distance between c & T: $\frac{|3(m)+5(-1)+4-5m|}{\sqrt{m^2+(-1)^2}}=\sqrt{5}\rightarrow\frac{(-2m-1)^2}{m^2+1}=5\rightarrow 4m^2+4m+1=5m^2+5\rightarrow m^2-4m+4\,=0\rightarrow (m-2)(m-2)=0\rightarrow m=2$ For $m=2\rightarrow 2x-y+4-5(2)=0\rightarrow2x-y-6=0$ Ans: (ii) (From text book): $2x+y-6=0$
 August 30th, 2013, 07:20 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: The Equation of a Tangent to a Circle If I we going to work part (ii), I would let the tangent line be: $y=m(x-1)+4$ Substituting this into the equation of the circle: $(x-3)^2+(m(x-1)-1)^2=5$ Arrange as quadratic in $x$: $x^2-6x+9+m^2(x-1)^2-2m(x-1)+1=5$ $x^2-6x+9+m^2$$x^2-2x+1$$-2mx+2m+1=5$ $x^2-6x+9+m^2x^2-2m^2x+m^2-2mx+2m+1=5$ $$$m^2\,+\,1)x^2\,-\,\(2m^2\,+\,2m\,+\,6$$x\,+\,m^2\,+\,2m\,+\,5\,=\, 0$ Equate the discriminant to zero since there is only 1 real root (repeated): $$$2m^2+2m+6$$^2-4$$m^2+1$$$$m^2+2m+5$$=0$ The reduces to: $(m+2)^2=0$ And so $m=-2$ and the tangent line is: $y=-2(x-1)+4$ $2x+y-6=0$ This is what your textbook has. Your error came from using the wrong point in the point-slope formula at the beginning of your working. You used the point (5,4) instead of (1,4).
 August 30th, 2013, 07:48 AM #3 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: The Equation of a Tangent to a Circle Great. Thank you very much.

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