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 March 31st, 2018, 05:59 AM #1 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [ASK] Tangent of a Circle One of the tangent line equation of the circle $\displaystyle x^2+y^2+6x-8y+12=0$ 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: $\displaystyle (-1)^2+y^2+6(-1)+8y+12=0$ $\displaystyle 1+y^2-6+8y+12=0$ $\displaystyle y^2+8y+7=0$ (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: 3,264 Thanks: 902 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: 714 Thanks: 151 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 $\displaystyle (x+3)^2+(y-4)^2=13$, you can find the two tangent lines by writing the circle this way: $\displaystyle (x+3)(x+3)+(y-4)(y-4)=13$ and substitute in the points: $\displaystyle (-1+3)(x+3)+(1-4)(y-4)=13$ to get $\displaystyle 2x-3y=-5$ and $\displaystyle (-1+3)(x+3)+(7-4)(y-4)=13$ to get $\displaystyle 2x+3y=19$ April 8th, 2018, 06:36 AM #4 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Yup, thanks. April 8th, 2018, 06:58 AM   #5
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Quote:
 Originally Posted by Country Boy . . . 3y- 21= -2x- 3, 2x+ 3y= 18.
That should be 3y - 21 = -2x - 2, 2x + 3y = 19. 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 thehandofomega Geometry 3 April 23rd, 2015 05:30 AM Peter White Trigonometry 4 February 21st, 2013 11:28 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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