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If the distance of a point P from (6,0) ...

If the distance of a point P from (6,0) is twice its distance from the point (1,3) prove that the locus of P is a circle. Find its centre and radius.

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A circle is the locus of a point in a plane such that its distance from a fixed point in the plane is constant. Anologously, a sphere is the locus of a point in space such that its distance from a fixed point in space in constant. The fixed point is called the centre and the constant distance is called the radius of the circle/sphere. In anology with the equation of the circle |z-c|=a , the equation of a sphere of radius is |r-c|=a , where c is the position vector of the centre and r is the position vector of any point on the surface of the sphere. In Cartesian system, the equation of the sphere, with centre at (-g, -f, -h) is x^2+y^2+z^2+2gx+2fy+2hz+c=0 and its radius is sqrt(f^2+g^2+h^2-c) . Q. The centre of the sphere (x-4)(x+4)+(y-3)(y+3)+z^2=0 is

A circle is the locus of a point in a plane such that its distance from a fixed point in the plane is constant. Anologously, a sphere is the locus of a point in space such that its distance from a fixed point in space in constant. The fixed point is called the centre and the constant distance is called the radius of the circle/sphere. In anology with the equation of the circle |z-c|=a , the equation of a sphere of radius is |r-c|=a , where c is the position vector of the centre and r is the position vector of any point on the surface of the sphere. In Cartesian system, the equation of the sphere, with centre at (-g, -f, -h) is x^2+y^2+z^2+2gx+2fy+2hz+c=0 and its radius is sqrt(f^2+g^2+h^2-c) . Q. Radius of the sphere, with (2, -3, 4) and (-5, 6, -7) as xtremities of a diameter, is