Question

if the line by = root 3 X + K touches the circle X square + Y square = 16 then find the value of k​

Answer 1

Step-by-step explanation:

$Given \: it \: \\ circle \: {x}^{2} + {y}^{2} = 16 \\ radius \: of \: circle = 4 \: unit \\ centre = (0 \:, 0) \\ If \: the \: line \: y = \sqrt{3} x + k \: touches \: the \: circle \\ then \: perpendicular \: drawn \: from \: centre = radius \\ \: \: \: \: \: \: \: \: \: \: \: \frac{ - k}{ \sqrt{1 + 3} } = 4 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{ - k}{2} = 4 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: k = - 8 \: \\ hope \: it \: will \: help \: you.$

Answer 2

Here the circle has centre at (0, 0) and radius 4.

The line touches the circle at a point. So, the distance between this point and the centre is equal to the radius of the circle.

This distance is the same as the perpendicular distance between the centre and the line.

We know that the perpendicular distance from the point (x1,y1) to the line ax + by + c = 0 is given by