My Math Forum [ASK] Tangent of a Circle

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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.

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