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

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

## The correct option is D (x−p)2=4qyIn a circle, AB is a diameter where the coordintes of A are (p, q). Let the coordinates of B be (x1,y1). The equation of the circle in diameter form is (x−p)(x−x1)+(y−q)(y−y1)=0⇒x2−(p+x1)x+px1+y2−(y1+q)y+qy1=0⇒x2−(p+x1)x+y2−(y1+q)y+px1+qy1=0 Since this circle touches the x-axis, ∴ y=0 ⇒x2−(p+x1)x+px1+qy1=0 Also the discriminant of the above equation will be equal to zero because the circle touches the x-axis. Therefore, (p+x1)2=4(px1+qy1)⇒p2+x21+2px1=4px1+4qy1⇒x21−2px1+p2=4qy1 Therefore, the locus of the point B is (x−p)2=4qy

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