Question

A variable circle passes through the fixed point A(p,q) and touches x-axis. The locus of the other end of the diameter through A 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 $(x-p{)}^2=4qy$

From Image $1$

Let the center of the circle be at $(a,b)$

It passes through $(a,0)$ and radius will be 'b'

Equation of circle will be

${(x-a)}^2+{(y-b)}^2=b^2$

(Image 2)

$x=2a-py=2b-q$

$a=\frac{x+p}{2}b=\frac{y+q}{2}$

The circle should pass through $(x,y)$
Substituting $(a,b)$ in ${(x-a)}^2+{(y-b)}^2=b^2$

${(x-\frac{x+p}{2})}^2+{(y-\frac{y+q}{2})}^2={\left(\frac{y+q}{2}\right)}^2$

$\frac{{(x-p)}^2}{4}+\frac{{(y-q)}^2}{4}=\frac{{(y+q)}^2}{4}$

${(x-p)}^2={(y+q)}^2-{(y-q)}^2$

$=y^2+q^2+2yq-y^2-q^2+2yq$

$\therefore {(x-p)}^2=4yq$