A triangle has two of its sides along the axes, its third side touches the circle x2+y2−2ax−2ay+a2=0, where a>0. If the locus of the circumcentre of the triangle passes through the point (38,−37), then a2−2a is equal to
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Solution
The given circle has its centre at C(a,a) and its radius is a, so that it touches both the axes along which lie the two sides of the triangle.
Let third side be xp+yq=1 So that A is (p,0) and B is (0,q) and the line AB touches the given circle. Since ∠AOB is a right angle, AB is a diameter of the circumcircle of the triangle AOB. So, the circumcentre P(h,k) of the triangle AOB is the mid point of AB. i.e. 2h=p,2k=q Now, the equation of AB is,xp+yq=1 which touches the given circle. Therefore, a(p+q)−pq√p2+q2=a ⇒a2(p+q)2+p2q2−2apq(p+q)=a,2(p2+q2) ⇒2a2−2a(p+q)+pq=0 ⇒2a2−2a(2h+2k)+2h⋅2k=0.
Hence, the locus of P(h,k) is a2−2a(x+y)+2xy=0. Since it passes through (38,−37) a2−2a=2×38×37=2812