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# the range of k

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If a straight line x-2y+k=0 does not touch the circle x^2+y^2-4x-6y+8=0,then the range of k is

Guest Feb 12, 2015

#2
+81029
+10

Here's the problem, algebraically.....

The slope of the line = (1/2)

And the slope at any point on the circle is given by

2x + 2yy' - 4 - 6y' = 0

y"(2y - 6) = (4 - 2x)

y' = (2 - x) / (y - 3)

And equating slopes, we have

(1/2) = (2 - x) / (y -3)     and these will be equal at (1, 5)

But, these slopes will also be equal when  (-1/-2) = (2 - x)/(y - 3) ...which gives the point, (3, 1)

So, when x = 1 and y = 5, we have 1 - 2(5) + k = 0  and k = 9

And when x = 3 and y = 1, we have 3 - 2(1) + k = 0  and k= -1

So, the lines will be tangent to the circle when k = -1 and k = 9. And this implies that  the line will intercept the circle for any k values on [-1, 9].

Therefore, the range of k values in which the given line does NOT touch the circle is given by  (-∞, -1) U (9, ∞)

CPhill  Feb 12, 2015
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#1
+81029
+5

I just played around with this one......here's the graph......

https://www.desmos.com/calculator/gaq9rk2oel

The line will not touch the circle when k lies on these two intervals (-∞, -1) U (9, ∞)

CPhill  Feb 12, 2015
#2
+81029
+10

Here's the problem, algebraically.....

The slope of the line = (1/2)

And the slope at any point on the circle is given by

2x + 2yy' - 4 - 6y' = 0

y"(2y - 6) = (4 - 2x)

y' = (2 - x) / (y - 3)

And equating slopes, we have

(1/2) = (2 - x) / (y -3)     and these will be equal at (1, 5)

But, these slopes will also be equal when  (-1/-2) = (2 - x)/(y - 3) ...which gives the point, (3, 1)

So, when x = 1 and y = 5, we have 1 - 2(5) + k = 0  and k = 9

And when x = 3 and y = 1, we have 3 - 2(1) + k = 0  and k= -1

So, the lines will be tangent to the circle when k = -1 and k = 9. And this implies that  the line will intercept the circle for any k values on [-1, 9].

Therefore, the range of k values in which the given line does NOT touch the circle is given by  (-∞, -1) U (9, ∞)

CPhill  Feb 12, 2015

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