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

## Comments

Sort by:

TopNewestthnks ,sir

Log in to reply

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

Log in to reply