My Math Forum [ASK] Tangent of a Circle (Again)

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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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 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.

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