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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 $$A$$ is


A
(yq)2=4px
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B
(xq)2=4py
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C
(yp)2=4qx
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D
(xp)2=4qy
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Solution

The correct option is D $${ \left( x-p \right)  }^{ 2 }=4qy$$
Let the variable circle be
$${ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0$$    ...(1)
$$\therefore { p }^{ 2 }+{ q }^{ 2 }+2gp+2fy+c=0$$   ...(2)
Circle (1) touches x-axis,
$$\therefore { g }^{ 2 }-c=0\Rightarrow c={ g }^{ 2 }$$ 
From (2) $${ p }^{ 2 }+{ q }^{ 2 }+2gp+2fq+{ g }^{ 2 }=0$$   ...(3)
Let the other end of diameter through $$(p,q)$$ be $$(h,k)$$, then
$$\displaystyle \frac { h+p }{ 2 } =-g$$ and $$\displaystyle \frac { k+q }{ 2 } =-f$$
Put in (3)
$$\displaystyle { p }^{ 2 }+{ q }^{ 2 }+2p\left( -\frac { h+p }{ 2 }  \right) +2q\left( -\frac { k+q }{ 2 }  \right) +{ \left( \frac { h+p }{ 2 }  \right)  }^{ 2 }=0\\ \Rightarrow { h }^{ 2 }+{ p }^{ 2 }-2hp-4kq=0$$
$$\therefore$$ Locus of $$(h,k)$$ is
$${ x }^{ 2 }+{ p }^{ 2 }-2xp-4yk=0\Rightarrow { \left( x-p \right)  }^{ 2 }=4qy$$

Mathematics

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