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# find the length of the chord joining the points in which the straight line, (x/a)+(y/b)=1, meets the circle x^2+y^2=r^2 9 years ago

Hi Suhas,

Consider (p,q) to be the midpoint of the chord.

So T=S1 is its equation.

which is px+qy = p2+q2---------(this eqn is the same as x/a+y/b=1)

So ap = bq = ??p2+q2.

Let the line intersect the cirlcea at (x1,y1) and (x2,y2)

So D = √[ (x1-x2)2+(y1-y2)2 ] = √[ (x1+x2)2+(y1+y2)2 - 4x1x2 - 4y1y2 ]

Now x1+x2 = 2p, y1+y2 = 2q --------(As (p,q) is the midpoint).

From the line equation px+qy = p2+q2, eliminate y from the circle eqn. You'll get a QE in x, for which x1,x2 are the roots.

So you'll get product of roots = x1x2 = p2+q2 - r2q2.

similarly y1y2 = p2+q2 - r2p2.

Substitute in the distance, and you'll get

D = √[4r2 - 4(p2+q2)]

Now ap=bq=p2+q2.

Also p/a+q/b=1

So p2+q2 = (a2b2)/(a2+b2)......

Now substitute and get D, interms of r,a,b

Best Regards,