Question

A circle cutting the circle $x^2+y^2=4$ orthogonally and having its centre on the line $2x-2y+9=0$ passes through two fixed points. These points are

• A. $(4,0)$ and $(0,4)$
• B. $(-4,4)$ and $(\frac{-1}{2},\frac{1}{2})$
• C. $(-4,0)$ and $(4,0)$
• D. $(4,-4)$ and $(\frac{1}{2},\frac{-1}{2})$

Answer 1

The correct option is B $(-4,4)$ and $(\frac{-1}{2},\frac{1}{2})$

Let the equation of circle be

$x^2+y^2+2yx+2+y+c=0$ __ (1)

$x^2+y^2-4=0$__(ii)

Equation (i) & (ii) are orthogonal

If $2g_1{g}_2+2l_1{l}_2=c_1+c_2$

$\Rightarrow 2g\times 0+2l\times 0=c-4$

$\Rightarrow c=4$ __ (iii)

since, (-g, -l) lies 2x - 2y +9 = 0

$\Rightarrow -2g+2l+9=0$

$\Rightarrow 2l=2g-9$ __(iv)

using (iii) & (iv) in (ii)

$x^2+y^2+2gx+(2g-9)y+4=0$

$\Rightarrow x^2+y^2+2gy-gy+4=0$

$\Rightarrow (x^2+y^2-gy+4)+2g(x+y)=0$

$S-1\lambda L=0$

$S:x^2+y^2-gy+y=0$

$L:x+y=0$

$\therefore x^2+x^2+9x+4=0$

$\therefore x=\frac{-1}{2},x=-4$

$\therefore y=\frac{1}{1},y=4$

$P(\frac{-1}{2},\frac{1}{2}),Q(-4,4)$