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  •  The area of the triangle formed by the pair of tangents from <img alt="\mathrm{\left(\mathrm{x}_1, \mathrm{y}_1\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

 The area of the triangle formed by the pair of tangents from \mathrm{\left(\mathrm{x}_1, \mathrm{y}_1\right)} to the circle \mathrm{x^2+y^2=a^2} and the chord joining their points of contact with the circle is \mathrm{\mathrm{k}\left(\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2\right)^{3 / 2}\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^{-1} \text { sq. units }} where k=

Option: 1

2a


Option: 2

\frac{a}{2}


Option: 3

a


Option: 4

2


Answers (1)

best_answer

Let \mathrm{R}\left(\mathrm{x}_1, \mathrm{y}_1\right)  be the point from which are drawn the tangents RP and RQ to the circle 2 \mathrm{k}\left(\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2\right)^{3 / 2}\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)^{-1} \text { sq. units }.

Let OR meet PQ at L, where O(0, 0) is the centre of the circle.

Then \Delta \mathrm{RPQ}=2 \Delta \mathrm{RLP}=2[\Delta \mathrm{OPR}-\Delta \mathrm{OLP}]

but \begin{aligned} & \Delta \mathrm{OPR}=1 / 2 \mathrm{OP} \cdot \mathrm{PR}=1 / 2 \mathrm{a} \cdot \sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2-\mathrm{a}^2} \\ & \Delta \mathrm{OLP}=1 / 2 \mathrm{OL} \cdot \mathrm{LP}=1 / 2(\mathrm{OP} \cos \theta)(\mathrm{OP} \sin \theta) \text { where } \angle \mathrm{POR}=\theta \end{aligned}

\Delta \mathrm{OPR}, \sin \theta=\frac{\mathrm{PR}}{\mathrm{OR}}=\frac{\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2-\mathrm{a}^2}}{\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2}} \text { and } \cos \theta=\frac{\mathrm{OP}}{\mathrm{OR}}=\frac{\mathrm{a}}{\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2}}

hence \Delta \mathrm{LOP}=1 / 2 \mathrm{a}^2 \cos \theta \sin \theta=                 \begin{aligned} & \frac{\sqrt{x_1^2+y_1^2-a^2}}{2\left(x_1^2+y_1^2\right)} a^3 \\ & \text { and } \triangle \mathrm{RPQ}=2\left[\frac{1}{2} a \sqrt{x_1^2+y_1^2-a^2}-\frac{1}{2} a^3 \frac{\sqrt{x_1^2+y_1^2-a^2}}{x_1^2+y_1^2}\right] \end{aligned}

=a \sqrt{x_1^2+y_1^2-a^2}\left[1-\frac{a^2}{x_1^2+y_1^2}\right]=a \sqrt{x_1^2+y_1^2-a^2} \frac{\left(x_1^2+y_1^2-a^2\right)}{x_1^2+y_1^2}=\frac{a\left(x_1^2+y_1^2-a^2\right)^{3 / 2}}{x_1^2+y_1^2} sq.unit

Posted by

Devendra Khairwa

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