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Question

A triangle has two of its sides along the axes, its third side touches the circle x2+y22ax2ay+a2=0.Prove that the locus of the circum-centre of the triangle is a22a(x+y)+2xy=0.

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Solution

The given circle has its centre (a,a) and radius a so that this circle touches both the axes along which lie the two sides of the triangles.
Let the third side be x/p+y/q=1,
so that P is (p,0) and Q is (0,q). The line PQ touches the given circle at the point R.
Since POQ is a right angle therefore PQ is diameter of the circle passing through the points O,P and Q and its centre is mid-point R of PQ.
R is (p/2,q/2)=(h,k), say.
p=2h,q=2k
Now the line PQ is x/p+y/q=1
or qx+pypq=0
touches the given circle. p=r
or qa+papq(p2+q2)=a. Square
a2(p+q)2+p2q22pqa(p+q)=a2(p2+q2)
or a2.2pq+p2q22pqa(p+q)=0 cancel pq
or 2a22a(p+q)+pq=0
or 2a22a(2h+2k)+2h.2k=0
Locus of centre (h,k) is
a22a(x+y)+2xy=0.
924028_1007998_ans_dae5c8af4160474aa990388360a78d82.png

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