an excellent prob of circles !! JUST SOLVE IT!!

the area of the triangle formed by the tangents from the points (h,k) to the circle x^2 +y^2=a^2 and the line joining the pint of contact is-----

Easy Math Editor

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Comments

If $P$ is the point $(h,k)$ , $X,Y$ the points of tangency with the circle, and $O$ the centre of the circle, then the kite $OXPY$ has area $a\sqrt{h^2+k^2-a^2}$ , while the triangle $OXY$ has area $a^2\sin\theta\cos\theta$ , where $\theta = \angle XOP$ . Since $\sin\theta \; = \; \frac{\sqrt{h^2+k^2-a^2}}{\sqrt{h^2+k^2}} \qquad \cos\theta \; = \; \frac{a}{\sqrt{h^2+k^2}}$ we deduce that the area of the triangle $PXY$ is $\frac{a(h^2+k^2-a^2)^{\frac32}}{h^2+k^2}$

Mark Hennings - 7 years, 7 months ago

thnks ,sir

sumedh bang - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$