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-----

Note by Sumedh Bang
4 years, 8 months ago

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thnks ,sir

- 4 years, 8 months ago

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 $\frac{a(h^2+k^2-a^2)^{\frac32}}{h^2+k^2}$

- 4 years, 8 months ago