A variable circle passes through the fixed point A(p,q) and touches the x-axis.

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:

Accepted Answer

Let the variable circle be

$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ $${p^2} + {q^2} + 2gp + 2fq + c = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Circle $(1)$ touches $x$-axis,

$$\therefore$$ $${g^2} - c = 0 \Rightarrow c = {g^2}.\,\,\,$$

From $(2)$

$${p^2} + {q^2} + 2gp + 2fq + {g^2} = 0\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$

Let the other end of diameter through $(p,q)$ be $(h,k),$

then, $${{h + p} \over 2} =  - g$$ and $${{k + q} \over 2} =  - f$$

Put in $(3)$

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

$$ \Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0$$

$$\therefore$$ locus of $$\left( {h,k} \right)$$ is ${x^2} + {p^2} - 2xp - 4yq = 0$

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

Solutions · 1