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Question

A variable circle passes through the points $$P(1,2)$$ and touches the x-axis. The locus of the other end of the diameter through $$P$$ is


Solution

eqn of circle $$\rightarrow$$
$$(x-1)(x-h)+(y-2)(y-k)=0$$
$$x^2+y^2-(h+1)x-(b+2)y+h+2b=0$$
Since, the circle touches the $$x-axis$$
$$g^2=c$$
$$\left(\dfrac{h+1}{2}\right)^2=(h+2b)$$
$$(x+1)^2=4(x+2y)$$

1374974_1161155_ans_16bb4a0eb30f4a63aca11d38f4dbeab7.png

Mathematics

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