Multiple choice

If pair of tangents $PA$ and $PB$ are draw from point $P(2,3)$ to the circle ${x}^{2}+{y}^{2}+4x+2y+1=0$ area $PAOB$, $O$ is centre.

  1. $2\sqrt{7}$
  2. $\sqrt{7}$
  3. $4\sqrt{7}$
  4. $8\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Circle: x^2 + y^2 + 4x + 2y + 1 = 0. Center O = (-2, -1). Radius r = sqrt(2^2 + 1^2 - 1) = 2. Point P = (2, 3). Distance OP = sqrt((2 - (-2))^2 + (3 - (-1))^2) = sqrt(4^2 + 4^2) = sqrt(32) = 4 * sqrt(2). Length of tangent PA = sqrt(OP^2 - r^2) = sqrt(32 - 4) = sqrt(28) = 2 * sqrt(7). Area of quadrilateral PAOB = 2 * Area(triangle OAP) = 2 * (1/2 * base * height) = PA * r = 2 * sqrt(7) * 2 = 4 * sqrt(7).

AI explanation

The centre of the circle is O(-2, -1) and its radius is sqrt(4 + 1 - 1) = 2. The distance from P(2,3) to O(-2,-1) is sqrt(16 + 16) = sqrt(32). The length of tangent PA is sqrt(PO^2 - r^2) = sqrt(32 - 4) = sqrt(28). Area of PAOB equals 2 times area of triangle POA, which is 2 * (1/2) * OA * PA = 2 * 2 * sqrt(28) = 4 * sqrt(7).

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