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
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, ∞)
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, ∞)
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, ∞)