March 31st, 2018, 05:59 AM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry  [ASK] Tangent of a Circle
One of the tangent line equation of the circle $\displaystyle x^2+y^2+6x8y+12=0$ at the point whose abscissa is 1 is .... A. 2x  3y  7 = 0 B. 2x  3y + 7 = 0 C. 2x + 3y  5 = 0 D. 2x  3y  5 = 0 E. 2x  3y + 5 = 0 By substituting x = 1, I got: $\displaystyle (1)^2+y^2+6(1)+8y+12=0$ $\displaystyle 1+y^26+8y+12=0$ $\displaystyle y^2+8y+7=0$ (y + 1) (y + 7) = 0 y = 1 or y = 7 Then what? Last edited by skipjack; April 8th, 2018 at 06:51 AM. 
March 31st, 2018, 12:06 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Completing the squares, $\displaystyle x^2+ 6x+ 9+ y^2 8x+ 16+ 12 9 16= (x+ 3)^2+ (y 4)^2 13= 0$ so this circle has center (3, 4) and has radius $\displaystyle \sqrt{13}$. If x= 1, yes, $\displaystyle (1+ 3)^2+ (y 4)^2 13= (y 4)^2 9= 0$, $\displaystyle y 4= \pm 3$ so y= 7 or y= 1. The "abscissa" x= 1 crosses the given circle at (1, 1) and (1, 7) so there are two such tangent lines and two possible answers though the question only asks for "one of" them. The line through the center (3, 4) to (1, 1) has slope (1 4)/(1 (3))= 3/2 so the tangent line, which is perpendicular to the radius, has slope 2/3. The line through (1, 1) with slope 2/3 is y 1= (2/3)(x (1)). y 1= (2/3)x+ 2/3, or 3y 3= 2x+ 2 , 2x 3y= 5. The line through the center (3, 4) to (1, 7) has slope (7 4)/(1(3))= 3/2. The tangent line has slope 2/3. The line through (1, 7) with slope 2/3 is y 7= (2/3)(x+ 1), 3y 21= 2x 3, 2x+ 3y= 18. 
April 3rd, 2018, 05:58 AM  #3 
Senior Member Joined: Feb 2010 Posts: 694 Thanks: 132 
Once you have substituted in $\displaystyle x=1$ to get the two points on the circle $\displaystyle (1,1)$ and $\displaystyle (1,7)$ and also completed the square to get $\displaystyle (x+3)^2+(y4)^2=13$, you can find the two tangent lines by writing the circle this way: $\displaystyle (x+3)(x+3)+(y4)(y4)=13$ and substitute in the points: $\displaystyle (1+3)(x+3)+(14)(y4)=13$ to get $\displaystyle 2x3y=5$ and $\displaystyle (1+3)(x+3)+(74)(y4)=13$ to get $\displaystyle 2x+3y=19$ 
April 8th, 2018, 06:36 AM  #4 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
Yup, thanks.

April 8th, 2018, 06:58 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,723 Thanks: 1807  

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