Question

A variable circle passes through the fixed point A(p, q) and touches x-axis. If locus of the other end of the diameter through A is $(x-p{)}^2=\lambda qy$, then the value of $\lambda$ is

• A. 4
• B. 5
• C. 6
• D. 9

Answer 1

The correct option is A 4

$(x-p)(x-h)+(y-q)(y-k)=0$
$\Rightarrow x^2+y^2-x(h+p)-y(q+k)+(ph+qk)=0$
This circle touches x-axis $g^2=c$
$\Rightarrow {\left(\frac{h+p}{2}\right)}^2=ph+qk$
Locus of (h, k) is $(x-p{)}^2=4qy$