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y=mx is a chord of the circle of radius `a` through the origin & whose diameter is along the axis of x. Find the equation of the circles whose diameter is the chord. Hence find the locus of its center for all values of m.
Since circle passes through the origin & diameter along the x-axis therefore its centre is (a,0) and its eq. is (x-a)2+y2=a2or x2+y2-2ax=0.............1equation of the circle passing through the circle(1) & y=mxx2+y2-2ax+c(y-mx)=0 where c is any constant its center is(a+m/2,-c/2)center lies on y=mx-c/2=m(a+m/2)putting the value of c in eq be get the required answer.Thanks & RegardsSunil RaikwaraskIITian's Faculty
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