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

  • Option 1)

    (y-p)^{2}=4qx

  • Option 2)

    (x-q)^{2}=4py

  • Option 3)

    (x-p)^{2}=4qy

  • Option 4)

    (y-q)^{2}=4px

 

Answers (1)

best_answer

As we learnt in

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 

Circle touching x-axis and having radius r -

x^{2}+y^{2}\pm 2rx+2fy+f^{2}= 0

- wherein

Where f is a variable parameter.

 

 Let the other diametric end be P(h,k)

So centre is \left (\frac{p+h}{2},\frac{q+R}{2} \right )

Radius =\sqrt{\left(\frac{h-p}{2} \right )^{2}+\left(\frac{k-q}{2} \right )^{2}}

For circle touching x-axis, radius =\left(\frac{q+k}{2} \right )

So\; \left(\frac{h-p}{2} \right )^{2}+\left(\frac{k-q}{2} \right )^{2}=\left(\frac{k+q}{2} \right )^{2}

we get (h-p)2=4kg 

i.e. (x-p)2=4qy. a parabola 

 


Option 1)

(y-p)^{2}=4qx

This option is incorrect 

Option 2)

(x-q)^{2}=4py

This option is incorrect 

Option 3)

(x-p)^{2}=4qy

This option is correct 

Option 4)

(y-q)^{2}=4px

This option is incorrect 

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