If a circle passes through the point (a,b) and cuts the circle x^{2}+y^{2}=4 orthogonally, then the locus of its centre is

  • Option 1)

    2ax-2by+(a^{2}+b^{2}+4)=0

  • Option 2)

    2ax+2by-(a^{2}+b^{2}+4)=0

  • Option 3)

    2ax+2by+(a^{2}+b^{2}+4)=0

  • Option 4)

    2ax-2by-(a^{2}+b^{2}+4)=0

 

Answers (1)

As we learnt in

Orthogonality of two circle -

Two circles S_{1}=0 and S_{2}=0  are said to be orthogonal ,if tangents at their point of intersection include right angle.

 

- wherein

2g_{1}g_{2}+2f_{1}f_{2}=c_{1}c_{2}

 

 x^{2}+y^{2}=4;

Circle pass through (a,b)

Let the variable circle be 

x2+y2+2gx+2fy+c=0                                    .........(1)

It passes through (a,b)

a2+b2+2ag+2fb+c=0

also  (1) cuts x2+y2=4 orthogonally 

So, 2(g1g2+f1f2)=C1+C2

2(0+0)=C_{1}-4

\Rightarrow    C1=4

Thus 2ax+2by=a2+b2-4


Option 1)

2ax-2by+(a^{2}+b^{2}+4)=0

This option is incorrect 

Option 2)

2ax+2by-(a^{2}+b^{2}+4)=0

This option is correct 

Option 3)

2ax+2by+(a^{2}+b^{2}+4)=0

This option is incorrect 

Option 4)

2ax-2by-(a^{2}+b^{2}+4)=0

This option is incorrect 

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