Circle O is centered at the point of origin with point P=(3,4) lying on it. The red line l : 3x + 4y - 7 = 0 intersects the circle at points A and B, as shown. What is the area of quadrilateral AOBP?
+128406 The radius of the circle = 5
The equation of the circle is x^2 + y^2 = 25 (1)
Re-write the equation of the line as y =(-3/4)x + 7/4 (2)
Sub (2) into (1)
x^2 + ( 7/4 -(3/4)x)^2 = 25
x^2 + (1/16)(3x - 7)^2 = 25
16x^2 + 9x^2 - 42x + 49 = 400
25x^2 - 42x -351 = 0
Solving this for x gives x = -3 and x = 117/25
And we only needt to find one associated value for y because AP = BP
So A = (-3 , (-3/4)(-3) + 7/4) = (-3, 4)
And AP = sqrt [ ( -3 - 3)^2 + (4 - 4)^2 ] = sqrt [36] = 6
And OP = 5 and OA = 5
So semi-perimeter of OAP = [ 5 + 5 + 6] / 2 = 8
So....using symmetry [AOBP ] =
2sqrt [ 8 * ( 8 - 5)^2 * (8 - 6) ] =
2 sqrt [ 8 * 9 * 2 ] =
2sqrt [ 144 ] =
2 * 12 =
24
