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

  • Option 1)

    2ax+2by-(a^{2}-b^{2}+p^{2})=0\;

  • Option 2)

    x^{2}+y^{2}-3ax-4by+(a^{2}+b^{2}-p^{2})=0

  • Option 3)

    2ax+2by-(a^{2}+b^{2}+p^{2})=0\;

  • Option 4)

    x^{2}+y^{2}-2ax-3by+(a^{2}-b^{2}-p^{2})=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}

 

 Let the variable circle is    

x2 + y2 + 2gx + 2fy + C = 0

It passes through ( a, b)

a2+b2+2ga+2fb+c=0                                                     -- (i)

It cuts x2 + y2 =p2 orthogonally.

2(g\times 0+f \times 0)=c-p^{2}

so, c = p2

For locus of (-g, -f), Replace it by (x, y) in (i)

-2ax - 2by + (a2 + b2 + p2) = 0

so, 2ax + 2by - (a2 + b2 + p2) = 0


Option 1)

2ax+2by-(a^{2}-b^{2}+p^{2})=0\;

This is incorrect option

Option 2)

x^{2}+y^{2}-3ax-4by+(a^{2}+b^{2}-p^{2})=0

This is incorrect option

Option 3)

2ax+2by-(a^{2}+b^{2}+p^{2})=0\;

This is correct option

Option 4)

x^{2}+y^{2}-2ax-3by+(a^{2}-b^{2}-p^{2})=0

This is incorrect option

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