Two tangents PQ and PR are drawn to the circl | vedclass

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(A) The area of the quadrilateral $PQCR$ is the sum of the areas of two congruent right-angled triangles,$\Delta PQC$ and $\Delta PRC$.Area $(PQCR) =

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The area of the quadrilateral $PQCR$ is the sum of the areas of two congruent right-angled triangles,$\Delta PQC$ and $\Delta PRC$.Area $(PQCR) = 2 	imes 	ext{Area}(\Delta PQC) = 2 	imes (\frac{1}{2} 	imes PQ 	imes QC) = PQ 	imes r$.Here,$r$ is the radius of the circle and $PQ$ is the length of the tangent from point $P$.The equation of the circle is $x^2 + y^2 - 2x - 4y - 20 = 0$.The centre $C$ is $(1, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-2)^2 - (-20)} = \sqrt{1 + 4 + 20} = \sqrt{25} = 5$.The length of the tangent $L = PQ = \sqrt{S_1}$,where $S_1$ is the value of the circle equation at point $P(16, 7)$.$PQ = \sqrt{16^2 + 7^2 - 2(16) - 4(7) - 20} = \sqrt{256 + 49 - 32 - 28 - 20} = \sqrt{225} = 15$.Therefore,the area of quadrilateral $PQCR = PQ 	imes r = 15 	imes 5 = 75 	ext{ sq. units}$.

Two tangents $PQ$ and $PR$ are drawn to the circle $x^2 + y^2 - 2x - 4y - 20 = 0$ from the point $P(16, 7)$. If the centre of the circle is $C$,then the area of quadrilateral $PQCR$ is ............ $sq. \text{ units}$.

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