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A line tangent to a circle

Question:

Line L1 is tangent to circle x2 + y2 = 4 in quadrant II and perpendicular to line L2 x + y - 2 = 0. Find the equation of the line L1.

Solution:

line tangent to a circle and line perpendicular to another line.

Line L1 is tangent to the circle x2 + y2 = 4. P is the point of tangency. Connect OP. PO is perpendicular to L1 at point P. OP is equal to the radius of the circle which is 2. That is, OP = r = 2.

From the circle equation x2 + y2 = 4, the center of the circle is (0, 0) and the radius of the circle is r = 2.

L1 has the equation y = k1x + b, in which k1 is the slope of L1 and b is the y intercept of L1. Because L1 is perpendicular to L2, so k1 = - 1/k2. Now, we will find k2

Because L2 has the equation: x + y - 2 = 0, which change to y = -x + 2, so k2 = -1.

k1 = - 1/k2 = - [1/(-1)] = 1
so, L1 has the equation: y = x + b
we need a point in line L1 to determine the value of b
we use the formular:
d = |Ax0 + By0 + C|/square root of (A2 + B2)

In which d is the distance from a point to a line. The point had the coordinate O(x0, y0) and the line has the equation: Ax + By + C = 0.

Because the line is: x - y + b = 0, so, A = 1, B = -1 and C = b. The distance from the center of the circle to the line L1 is 2. Substitute these values into the formula.

|b| ÷ square root of [12 + (-1)2] = 2.
|b| = 2 (square root of 2)
line tangent to a circle and perpendicular to another circle.

For line L1, y-intercept is a positive number, so b = 2 (square root of 2). So, the line L1 has the equation: y = x + 2 (square root of 2). Watch the video for more details.