Let $A(2,3), B(3,-6), C(5,-7)$ be three points. If $P$ is a point satisfying the condition $PA^2+PB^2=2PC^2$,then a point that lies on the locus of $P$ is

If the ends of the hypotenuse of a right-angled triangle are $(0, a)$ and $(a, 0)$,then the locus of the third vertex is:

$A$ particle is projected vertically upwards. If it has to stay above the ground for $12 \text{ seconds}$,then:

$A$ point on the straight line $3x + 5y = 15$ which is equidistant from the coordinate axes will lie only in

The equation of the locus of the point of intersection of the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ is

If $A=(5,3)$,$B=(3,-2)$ and a point $P$ is such that the area of the triangle $PAB$ is $9$,then the locus of $P$ represents