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Deterimine whether the line 5x+12y=169 is a tangent to the circl  x2 + y2=169. if so, find the point where the tangent line touches the circle.

 Feb 6, 2017
 #1
avatar+130491 
0

 x^2 + y^2=169   (1)

5x+12y=169     (2)

 

The slope of any tangent line to the circle will be  =  -x /y

 

The slope of the line  =   -5/12

 

Equating slopes, we have

 

-x /y  = -5/12    →  y = 12/5 x       sub this into (1)  

 

x^2 +[ (12/5)x]^2  = 169

 

x^2 + (144/ 25)x^2  = 169

 

169x^2 / 25  = 169

 

x ^2 / 25   = 1

 

x^2  =25

 

x =  ± 5

 

When x = 5, y= (12/5) *5   = 12

 

When x  = -5, y  = (12/5) * -5  = - 12

 

Only   (5,12)   will  saitsfy  5x+12y=169

 

Here's a graph :

 

 

cool cool cool

 

 

 

 

 

 

 

x =

 Feb 7, 2017
 #2
avatar+26397 
+5

Deterimine whether the line 5x+12y=169 is a tangent to the circl  x2 + y2=169. if so, find the point where the tangent line touches the circle.

 

Discussion:
If the line and the circle have no intersections, then the line is not a tangent.
If the line and the circle have one intersection, then the line is a tangent.
If the line and the circle have two intersections, then the line is not a tangent.

 

We compute the intersections:

(1)5x+12y=169|169=1325x+12y=13212y=1325xy=1325x12(2)x2+y2=169|169=132x2+y2=132|y=1325x12x2+(1325x12)2=132x2+(1325x)2122=132|122122x2+(1325x)2=122132122x2+13213221325x+52x2=122132122x2+52x2+13213210132x=122132(122+52)x2+13213210132x=122132|122+52=132132x2+13213210132x=122132|:132x2+13210x=122|122x210x+132122=0|132122=52x210x+52=0x210x+25=0x=10±1004252x=10±1001002x=10±02x=102x=5y=1325x12y=1325212|13252=122y=12212y=12

 

We have only one intersection at x = 5 and y = 12,

so the line touches the circle.

 

laugh

 Feb 7, 2017

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