MHT CET202526 Apr 2025Evening ShiftMathematicsCircleActual
Two tangents to the circle x^2+y^2=4 at the points A and B meet at P (-4,0) . Then the area of quadrilateral PAOB , where 'O' is the origin is
Options
- A8 3 sq. units
- B4 3 sq. units
- C4 3 sq. units
- D3 sq. units
Correct answer
C. 4 3 sq. units
Step-by-step solution
The circle x^2 + y^2 = 4 has center O(0,0) and radius 2 units. The external point is P(-4,0) with OP = 4 units. Tangents from P touch the circle at A and B , forming right angles with radii OA and OB . The quadrilateral PAOB consists of two congruent right triangles. Using the Pythagorean theorem in OAP : PA^2 = OP^2 - OA^2 = 16 - 4 = 12 , so PA = 2 3 units. The area of OAP is 1 2 OA PA = 1 2 2 2 3 = 2 3 sq. units. Area of quadrilateral PAOB = 2 2 3 = 4 3 sq. units.