Let L and M be two perpendicular lines tangent to a circle with radius 6. Find the area bounded by the two lines and the circle
I see it as
the area of a 6 by 6 square - (1/4)circle
= 36 - (1/4)(π)6^2
= 36 - 9π
= appr. 7.73
Subtract the area of the circle from the 6x6 square. Divide the remaining square by four and add it (a quarter of the remaining) to the area of the circle.
To find the area bounded by the two lines and the circle, we need to determine the coordinates of the points of intersection and then calculate the individual areas of the sectors and triangles formed.
Let's start by finding the coordinates of the points of intersection between the lines L and M and the circle. Since the lines are perpendicular to each other and tangent to the circle, they will intersect with the circle at right angles.
Let's assume the center of the circle is at the origin (0,0). Since the radius of the circle is given as 6, the equation of the circle can be written as x^2 + y^2 = 36.
The equation of line L can be written in the form y = mx + b, where m is the slope and b is the y-intercept. Since the line is tangent to the circle, the slope of line L will be the negative reciprocal of the slope of the radius at the point of tangency. Since the radius is perpendicular to the tangent line, its slope is the negative reciprocal of the line M's slope.
Let's assume the equation of line M is x = a, where a is the x-coordinate of the point where M intersects the circle.
Now, let's find the equation of line L using the negative reciprocal of M's slope:
The slope of M is undefined since it is a vertical line. The negative reciprocal of an undefined slope is 0, so we can write the equation of line L as y = 0.
Now we have the equations x^2 + y^2 = 36 and y = 0. Substituting y = 0 into the equation of the circle, we get x^2 + 0^2 = 36, which simplifies to x^2 = 36. Taking the square root of both sides, we have x = ±6.
Since line M is perpendicular to line L, it will be a vertical line passing through one of the points of intersection (±6, 0). Let's assume it passes through the point (6, 0), which means a = 6.
Now we have the coordinates of the points of intersection: (6, 0), (-6, 0), and (0, 0).
To calculate the area bounded by the two lines and the circle, we need to break it down into sectors and triangles.
Since the lines are perpendicular and tangent to the circle, they divide the area into four equal sectors. The area of one sector can be calculated using the formula A = (θ/360) * π * r^2, where θ is the central angle of the sector and r is the radius of the circle.
In this case, the central angle of each sector is 90 degrees, so the area of one sector is (90/360) * π * 6^2 = 9π square units.
The area of the four sectors is 4 times the area of one sector, so the total area of the sectors is 4 * 9π = 36π square units.
Since the lines divide the circle into four congruent right-angled triangles, we can calculate the area of one triangle using the formula A = (1/2) * base * height.
The base of each triangle is 6 units (since it is the radius of the circle) and the height is also 6 units (since the line passes through the center of the circle).
Therefore, the area of one triangle is (1/2) * 6 * 6 = 18 square units.
The area of the four triangles is 4 times the area of one triangle, so the total area of the triangles is 4 * 18 = 72 square units.
Finally, to find the total area bounded by the two lines and the circle, we add the areas of the sectors and triangles together: 36π + 72 square units.
Therefore, the area bounded by the two lines and the circle is 36π + 72 square units.