Question

Which of the following statements (s) is/are correct with respect to the circles $S_1=x^2+y^2-4=0$ and $S_2=x^2+y^2-2x-4y+4=0$?

• A. $S_1$ and $S_2$ intersect at an angle of ${90}^o$
• B. The point of intersection of the two circle are $(2,0)$ and $(\frac{6}{5},\frac{8}{5})$.
• C. Length of the common of chord of $S_1$ and $S_2$ is $\frac{4}{\sqrt{5}}$.
• D. The point $(2,3)$ lies outside the circles $S_1$ and $S_2$.

Answer 1

The correct option is A $S_1$ and $S_2$ intersect at an angle of ${90}^o$

$S_1:x^2+y^2-4=0$

$S_2{x}^2+y^2-2x-4y+4=0$

$\therefore g=0$ $f=0$ $c=-4$

$g_1=-1$ $f_1=-2$ $c_1=4$

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

$\therefore 2(0+0)=4-4$

$\therefore 0=0$

Since above is zero $S_1$ and $S_2$ are orthogonal.