Question

The locus of the centres of all circle passing through $(1,2)$ and cutting $x^2+y^2=16$ orthogonally is

• A. $4x-2y+21=0$
• B. $x+2y+11=0$
• C. $2x+4y-21=0$
• D. $2x-y+11=0$

Answer 1

The correct option is C $2x+4y-21=0$

Let $(x_1,y_1)$ be the centre of circle

Also Equation of circle be

$x^2+y^2+2gx+2fy+c=0$

$\therefore$ Substituting $(1,2)$

$\therefore 1^2+2^2+2g\left(1\right)+2f\left(2\right)+c=0$

$\therefore 5+c=-2(g+2f)$

$\therefore g+2f=-\left(\frac{5+c}{2}\right)$

$\therefore 2g{g}_1+2f{f}_1=c+c_1⟶$ orthogonality condn

$\therefore 0+0=-16+c\to (g_1=f_1=0)$

$\therefore c=16$

$\therefore g+2f=-\left(\frac{5+16}{2}\right)$

$\therefore g+2f=\frac{-21}{2}$

Now $x_1=-g{y}_1=-f$

$\therefore -x_1-2y_1=\frac{-21}{2}$

$\therefore -2x_1-4y_1=-21$

$\therefore 2x_1+4y_1+21=0$

$\therefore$ Locus :- $2x+4y-21=0$