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 April 2nd, 2018, 06:48 AM #1 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [ASK] Tangent of a Circle (Again) If the line x + my = 1 is a tangent of the circle $\displaystyle x^2+y^2-4x+6y+8=0$, the value of m is .... A. -2 B. $\displaystyle \frac{1}{4}$ C. $\displaystyle \frac{1}{4}$ D. 3 E. 4 Looking at the circle's equation, the center is (2, -3) and the radius is $\displaystyle \sqrt5$. If I know the coordinate where the line meet the circle I think I can solve this by myself, but I don't know how. Should I substitute x = 1 - my to the circle equation, or is there a simpler way? April 2nd, 2018, 01:01 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I can't imagine why you would set x= 1. Do you have any reason to think x= 1 where the line intersects the circle? x+ my= 1 is equivalent to x= 1- my. So $\displaystyle x^2+ y^2- 4x+6y+ 8= (1- my)^2+ y^2- 4(1- my)+ 6y+ 8= 1- 2my+ m^2y^2+ y^2- 4+ 4my- 5y+ 8= (m^2+ 1)y^2+ (-2m+ 4m- 5)y+ (1- 4+ 8 )= (m^2+ 1)y^2+ (2m- 5)y+ 5= 0$. Solve that quadratic equation for y. April 2nd, 2018, 04:54 PM   #3
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Quote:
 Originally Posted by Country Boy I can't imagine why you would set x= 1. Do you have any reason to think x= 1 where the line intersects the circle? x+ my= 1 is equivalent to x= 1- my.
Where did I even say that I set x = 1? I said "Should I substitute x = 1 - my"? April 2nd, 2018, 06:31 PM   #4
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Quote:
 Originally Posted by Monox D. I-Fly If the line x + my = 1 is a tangent of the circle $\displaystyle x^2+y^2-4x+6y+8=0$, the value of m is .... A. -2 B. $\displaystyle \frac{1}{4}$ C. $\displaystyle \frac{1}{4}$ D. 3 E. 4
Since the answers are $\displaystyle 2$ and $\displaystyle \dfrac{-1}{2}$, I would say the answer is F. none of the above.

The radius is the distance from the center to any tangent line so

$\displaystyle \dfrac{\vert 2-3m-1 \vert}{\sqrt{1+m^2}}=\sqrt{5}$ or $\displaystyle (1-3m)^2 = 5(1+m^2)$, etc. April 2nd, 2018, 07:12 PM   #5
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 Originally Posted by Country Boy $\displaystyle (m^2+ 1)y^2+ (2m- 5)y+ 5= 0$. Solve that quadratic equation for y.
More easily, since the tangent touches but does not cross the circle, we need only find the values of $m$ for which the equation has two equal roots. That is where the discriminant $b^2-4ac =0$ or (since Country Boy has an error)
$$(2m+6)^2 = 4 \cdot 5(m^2+1)$$
This leads to $m \in \{-\frac12,2\}$ (as per mrtwhs).
Why are B and C the same? April 2nd, 2018, 07:37 PM #6 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Sorry, I was in a hurry so I did some typos. C was actually $\displaystyle \frac{1}{2}$ and D was actually 2. So, the answer is D. Tags circle, tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Monox D. I-Fly Geometry 4 April 8th, 2018 06:58 AM thehandofomega Geometry 3 April 23rd, 2015 05:30 AM snigdha Complex Analysis 5 April 17th, 2011 01:12 PM brian890 Trigonometry 3 March 4th, 2011 07:30 PM brian890 Trigonometry 1 February 25th, 2011 05:10 PM

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