Question

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 of the circle is _____

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

Answer 1

The correct option is B

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

Let's assume Centre of required circle is (-g, -f)

Equation of required circle be $x^2+y^2+2gx+2fy+c=0$ - - - - - - (1)

This circle passes through the point (a, b)

So, $a^2+b^2+2ga+2fb+c=0$

$c=-(a^2+b^2+2ga+2fb)$

Equation of required circle is

$x^2+y^2+2gx+2fy-(a^2+b^2+2ga+2fb)=0$ - - - - - - (2)

Required circle is orthogonal to $x^2+y^2-4=0$

$2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$2(-g)\left(0\right)+2(-f)\left(0\right)=-(a^2+b^2+2ga+2fb)-4$

$a^2+b^2+2ga+2fb+4=0$

$2ga+2fb+(a^2+b^2+4)=0$

To generalize the locus of center (-g, -f), we can replace g by (-x) and f by (-y)

Locus of the center of the circle is

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

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

Option B is correct