If a circle passes through the point $\dpi{100} (a,b)$ and cuts the circle $\dpi{100} x^{2}+y^{2}=4$ 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) $2ax-2by-(a^{2}+b^{2}+4)=0$

P Plabita

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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