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.
+130501 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 :
x =
+26398 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=132−5xy=132−5x12(2)x2+y2=169|169=132x2+y2=132|y=132−5x12x2+(132−5x12)2=132x2+(132−5x)2122=132|⋅122122x2+(132−5x)2=122⋅132122x2+132⋅132−2⋅132⋅5x+52x2=122⋅132122x2+52x2+132⋅132−10⋅132x=122⋅132(122+52)x2+132⋅132−10⋅132x=122⋅132|122+52=132132x2+132⋅132−10⋅132x=122⋅132|:132x2+132−10x=122|−122x2−10x+132−122=0|132−122=52x2−10x+52=0x2−10x+25=0x=10±√100−4⋅252x=10±√100−1002x=10±02x=102x=5y=132−5x12y=132−5212|132−52=122y=12212y=12
We have only one intersection at x = 5 and y = 12,
so the line touches the circle.
