Question

If the circles $(x+a{)}^2+(y+b{)}^2=a^2,(x+\alpha {)}^2+(y+\beta {)}^2={\beta }^2$
cut orthogonally then ${\alpha }^2+b^{2^{}}=$

• A. $a\alpha +b\beta$
• B. $a^2+{\beta }^2$
• C. $-2(a\alpha +b\beta )$
• D. $2(a\alpha +b\beta )$

Answer 1

Answer: (D)

$2(a\alpha +b\beta )$

If two circle cut orthogonally then,
$r_1^2+r_2^2=(c_1{c}_2{)}^2$
$\therefore (x+a{)}^2+(y+b{)}^2=a^2$
$r_1=a$ and $c_1=(-a,-b)$
and $(x+\alpha {)}^2+(y+\beta {)}^2={\beta }^2$
$r_2=\beta$ and $c_2=(-\alpha ,-\beta )$
$\Rightarrow a^2+{\beta }^2=a^2+{\alpha }^2=2\alpha a+{\beta }^2+b^2-2b\beta$
$\Rightarrow {\alpha }^2+{\beta }^2=2(\alpha a+\beta b)$