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.
+129840 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 =
+26387 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:
\(\begin{array}{|lrcll|} \hline (1) & 5x+12y &=& 169 \quad & | \quad 169 = 13^2 \\ & 5x+12y &=& 13^2 \\ & 12y &=& 13^2 -5x \\ & y &=& \frac{13^2 -5x}{12} \\\\ (2) & x^2 +y^2 &=& 169 \quad & | \quad 169 = 13^2 \\ & x^2 +y^2 &=& 13^2 \quad & | \quad y = \frac{13^2 -5x}{12} \\ & x^2 + (\frac{13^2 -5x}{12})^2 &=& 13^2 \\ & x^2 + \frac{(13^2 -5x)^2}{12^2} &=& 13^2 \quad & | \quad \cdot 12^2\\ & 12^2x^2 + (13^2 -5x)^2 &=& 12^2\cdot 13^2 \\ & 12^2x^2 + 13^2\cdot 13^2 -2\cdot 13^2\cdot 5x+5^2x^2 &=& 12^2\cdot 13^2 \\ & 12^2x^2 + 5^2x^2 + 13^2\cdot 13^2 -10\cdot 13^2x &=& 12^2\cdot 13^2 \\ & (12^2+5^2)x^2 + 13^2\cdot 13^2 -10\cdot 13^2x &=& 12^2\cdot 13^2 \quad & | \quad 12^2+5^2=13^2 \\ & 13^2x^2 + 13^2\cdot 13^2 -10\cdot 13^2x &=& 12^2\cdot 13^2 \quad & | \quad : 13^2 \\ & x^2 + 13^2 -10x &=& 12^2 \quad & | \quad - 12^2 \\ & x^2 -10x + 13^2 - 12^2 &=& 0 \quad & | \quad 13^2 - 12^2=5^2 \\\\ & x^2 -10x + 5^2 &=& 0 \\ & x^2 -10x + 25 &=& 0 \\\\ & x &=& \frac{10\pm \sqrt{100-4\cdot 25} }{2} \\ & x &=& \frac{10\pm \sqrt{100-100} }{2} \\ & x &=& \frac{10\pm 0 }{2} \\ & x &=& \frac{10}{2} \\ & \mathbf{ x } & \mathbf{=} & \mathbf{5} \\\\ & y &=& \frac{13^2 -5x}{12} \\ & y &=& \frac{13^2 -5^2}{12} \quad & | \quad 13^2 -5^2=12^2 \\ & y &=& \frac{12^2}{12} \\ & \mathbf{ y } & \mathbf{=} & \mathbf{12} \\ \hline \end{array} \)
We have only one intersection at x = 5 and y = 12,
so the line touches the circle.
