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 $\displaystyle x^2+y^24x+6y+8=0$, 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 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry  
April 2nd, 2018, 06:31 PM  #4  
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147  Quote:
The radius is the distance from the center to any tangent line so $\displaystyle \dfrac{\vert 23m1 \vert}{\sqrt{1+m^2}}=\sqrt{5}$ or $\displaystyle (13m)^2 = 5(1+m^2)$, etc.  
April 2nd, 2018, 07:12 PM  #5  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra  Quote:
$$(2m+6)^2 = 4 \cdot 5(m^2+1)$$ 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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