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From point P(4,0) tangents PA and PB are...

From point P(4,0) tangents PA and PB are drawn to the circle `S: x^2+y^2=4`. If point Q lies on the circle, then maximum area of `triangleQAB` is- (1) `2sqrt3` (2) `3sqrt3` (3) `4sqrt3` A) 9

A

12

B

27

C

48

D

45

Text Solution

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The correct Answer is:
To find the maximum area of triangle QAB where Q lies on the circle defined by the equation \( S: x^2 + y^2 = 4 \), we can follow these steps: ### Step 1: Understand the Geometry We have a point \( P(4, 0) \) from which tangents \( PA \) and \( PB \) are drawn to the circle \( S \). The center of the circle is at the origin \( O(0, 0) \) with a radius of \( r = 2 \). ### Step 2: Determine the Points of Tangency Using the properties of tangents, we can find the points \( A \) and \( B \) where the tangents touch the circle. The distance from point \( P \) to the center \( O \) is \( OP = 4 \) and the radius \( OA = OB = 2 \). ### Step 3: Use the Tangent Length Formula The length of the tangent from point \( P \) to the circle can be calculated using the formula: \[ PA = PB = \sqrt{OP^2 - OA^2} = \sqrt{4^2 - 2^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \] ### Step 4: Calculate the Area of Triangle QAB To maximize the area of triangle \( QAB \), we need to consider the height from point \( Q \) to line \( AB \). The maximum area occurs when \( Q \) is at the highest point on the circle, which is directly above the center \( O \) at point \( (0, 2) \). ### Step 5: Find the Length of AB Since \( A \) and \( B \) are the points of tangency, the length \( AB \) can be calculated as: \[ AB = 2 \cdot PA = 2 \cdot 2\sqrt{3} = 4\sqrt{3} \] ### Step 6: Calculate the Area The area \( A \) of triangle \( QAB \) can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Here, the base is \( AB \) and the height is the distance from point \( Q(0, 2) \) to line \( AB \) (which is the radius of the circle). Thus, the area becomes: \[ \text{Area} = \frac{1}{2} \times 4\sqrt{3} \times 2 = 4\sqrt{3} \] ### Conclusion The maximum area of triangle \( QAB \) is \( 4\sqrt{3} \). ### Final Answer The correct option is (3) \( 4\sqrt{3} \). ---
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