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  • Please,please help me - Co-ordinate geometry - JEE Main-6

The locus of the centres of the circles, which touch the circle,

x^{2}+y^{2}=1 externally , also touch the y-axis and lie in the

first quadrant, is :

  • Option 1)

    x=\sqrt{1+4y},y\geq 0

  • Option 2)

    y=\sqrt{1+2x},x\geq 0

  • Option 3)

    y=\sqrt{1+4x},x\geq 0

  • Option 4)

    x=\sqrt{1+2y},y\geq 0

Answers (1)

best_answer

Let the centre of one such circle be P(h,k) since it touches y-axis in the first quadrant.

=> its radius = h

Now, since it touches x^{2}+y^{2}=1

h+1=\sqrt{h^{2}+k^{2}}

=> h^{2}+2h+1=h^{2}+k^{2}

=> 2h+1=k^{2}

Hence the required locus is 

2x+1=y^{2}

\sqrt{2x+1}=y; x\geq 0

So, option (2) is correct


Option 1)

x=\sqrt{1+4y},y\geq 0

Option 2)

y=\sqrt{1+2x},x\geq 0

Option 3)

y=\sqrt{1+4x},x\geq 0

Option 4)

x=\sqrt{1+2y},y\geq 0

Posted by

Aadil

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