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A variable circle passes through the point A(a,b) and touches the x-axis. Show that the locus of the other end of the diameter through A is (x-a)^2=4by. Can anyone please help me in this question?

A variable circle passes through the point A(a,b) and touches the x-axis. Show that the locus of the other end of the diameter through A is (x-a)^2=4by.
Can anyone please help me in this question?

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1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
9 years ago
Let the other end of the diameter through A be (h,k)

So the coordinated of the center are (\frac{a+h}{2},\frac{b+k}{2})

Now you can calculate the radius of the circle as the coordinates of the centre and one point on the circle are known
Therefore, Radius = \sqrt{(\frac{h-a}{2})^2 + (\frac{k-b}{2})^2}

Now since the circle is touching the x-axis, radius=ordinate of center

\sqrt{(\frac{h-a}{2})^2 + (\frac{k-b}{2})^2} = \frac{k+b}{2}

Now simplify this and get the required.
Hence proved

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