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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
3 years, 7 months ago

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thnks ,sir Sumedh Bang · 3 years, 7 months ago

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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 · 3 years, 7 months ago

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