Question

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

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

Answer 1

The correct option is A

${\left(x-p\right)}^2=4qy$

Explanation for correct option :

Step-1: By using distance formula. $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

$\Rightarrow$$\left(\sqrt{{\left(h-\frac{h+p}{2}\right)}^2+{\left(k-\frac{k+q}{2}\right)}^2}\right)=\left(\sqrt{{\left(\frac{h+p}{2}-\frac{h+p}{2}\right)}^2+{\left(\frac{k+q}{2}-0\right)}^2}\right)$

Step-2: Squaring both sides.

$\Rightarrow$ ${\left(h-\frac{h+p}{2}\right)}^2+{\left(k-\frac{k+q}{2}\right)}^2={\left(\frac{k+q}{2}-0\right)}^2$

$\Rightarrow$ $\frac{{\left(h-p\right)}^2}{4}+\frac{{\left(k-q\right)}^2}{4}=\frac{{\left(k+q\right)}^2}{4}$

$\Rightarrow$ $h^2+p^2-2hp+k^2+q^2-2kq=k^2+q^2+2kq$

$\Rightarrow$ ${\left(h-p\right)}^2=4kq$

Step-3: Substitute $h=x\&k=y$

$\Rightarrow$ ${\left(x-p\right)}^2=4qy$

Hence, option $\left(A\right)$ is correct.