Question

A variable circle passes through the fixed point A(p, q) and touches the x-axis. The locus of the other end of the diameter through A is

• A. $(x-p{)}^2=4qy$
• B. $(x-q{)}^2=4py$
• C. $(y-p{)}^2=4qx$
• D. $(y-q{)}^2=4px$

Answer 1

The correct option is A $(x-p{)}^2=4qy$

In a circle, AB is a diameter where the coordintes of A are (p, q). Let the coordinates of B be $(x_1,y_1)$.
The equation of the circle in diameter form is
$(x-p)(x-x_1)+(y-q)(y-y_1)=0\Rightarrow x^2-(p+x_1)x+p{x}_1+y^2-(y_1+q)y+q{y}_1=0\Rightarrow x^2-(p+x_1)x+y^2-(y_1+q)y+p{x}_1+q{y}_1=0$
Since this circle touches the x-axis,
$\therefore$ y=0
$\Rightarrow x^2-(p+x_1)x+p{x}_1+q{y}_1=0$
Also the discriminant of the above equation will be equal to zero because the circle touches the x-axis. Therefore,
$(p+x_1{)}^2=4(p{x}_1+q{y}_1)\Rightarrow p^2+x_1^2+2p{x}_1=4p{x}_1+4q{y}_1\Rightarrow x_1^2-2p{x}_1+p^2=4q{y}_1$
Therefore, the locus of the point B is
$(x-p{)}^2=4qy$