First slide

Definition of a circle

Question

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

Easy

A

${(x - p)}^{2} = 4 q y$
B

${(x - q)}^{2} = 4 p y$
C

${(y - p)}^{2} = 4 q x$
D

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

Solution

Let $B (x_{1}, y_{1})$ be the other end of diameter.
$∴$ The equation of the circle with $A B$ as diameter is $(x - p) (x - x_{1}) + (y - y_{1}) (y - q) = 0$
$\Rightarrow x^{2} + y^{2} - x (p + x_{1}) - y (q + y_{1}) + p x_{1} + q y_{1} = 0$
$\Rightarrow 2 g = - (p + x_{1}) c = p x_{1} + q y_{1}$
Since circle touches x-axis
$\Rightarrow g^{2} = c$
$\Rightarrow \frac{{(p + x_{1})}^{2}}{4} = p x_{1} + q y_{1}$
$\Rightarrow {(x_{1} - p)}^{2} = 4 q y_{1}$
$∴$ locus is ${(x - p)}^{2} = 4 q y$

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