Question

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

• A. $x^2+y^2-3ax-4by+(a^2+b^2-p^2)=0$
• B. $2ax+2by-(a^2-b^2+p^2)=0$
• C. $x^2+y^2-2ax-3by+(a^2-b^2-p^2)=0$
• D. $2ax+3by-(a^2+b^2+p^2)=0$

Answer 1

The correct option is C $x^2+y^2-2ax-3by+(a^2-b^2-p^2)=0$

Let the equation of the circle which passes through the point (a,b) is given by,

$x^2+y^2+2gx+2fy+c=\mathrm{0..........}\left(1\right)$

$a^2+b^2+2ga+2fb+c=\mathrm{0........}\left(2\right)$

Now circle (1) cuts the circle $x^2+y^2=4$ orthogonally

Therefore,

$2g\times 0+2f\times 0=c-4$

$c=\mathrm{4.........}\left(3\right)$

Substituting $c=4$ in (2)

$a^2+b^2+2ga+2fb+4=\mathrm{0........}\left(2\right)$

Now let the centre of the circle by (h,k)

therefore $h=-g$ and $k=-f$

thus,

$a^2+b^2-2ha-2kb+4=0$

replace h by x and k by y,

thus the locus of the centre is

$a^2+b^2-2xa-2yb+4=0$

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