Question

A point P lies inside the circles $x^2+y^2-4=0$ and $x^2+y^2-8x+7=0$. The point P starts moving under the conditions that its path encloses greatest possible area and it is at a fixed distance from any arbitrarily chosen fixed point in its region. The locus of P is

Answer 1

$x^2+y^2=4$
$x^2+y^2-8x+7=0\Rightarrow (x-4{)}^2+y^2=(3{)}^2$
According to question point P lies in the region ABCD, and for it to cover maximum distance - the point has to travel along a circle with maximum diameter.
Hence with AB as diameter A(1,0) and B(2,0) - the mid-point is $[\frac{(1+2)}{2},\frac{0}{a}]=(1.5,0)$ or $(\frac{3}{2},0)$ with radius $=\frac{1}{2}$; the equation will be
${(x-\frac{3}{2})}^2+(y-0{)}^2={\left(\frac{1}{2}\right)}^2[\because (x-a{)}^2+(y-b{)}^2=r^2]\Rightarrow x^2-3x+\frac{9}{4}+y^2=\frac{1}{4}\Rightarrow x^2+y^2-3x+2=0$.