=1 Let P(5cosθ,4sinθ) be a point on 16x2+25y2=400. The equation of the tangent at P is 4xcosθ+5ysinθ=20 This meets the coordinate axes at A(5secθ,0) and A′(0,4cosecθ) The equation of the circle with AA′ as diameter is
(x−5secθ)(x−0)+(y−0)(y−4cosecθ)=0
⇒x2+y2−5xsecθ−4ycosecθ=0 Clearly, it passes through (0,0).