Question

The two circles $x^2+y^2-2x+22y+5=0$ and $x^2+y^2+14x+6y+k=0$ intersect orthogonally provided $k$ is equal to ?

• A. $47$
• B. $-47$
• C. $49$
• D. $-49$

Answer 1

The correct option is A $47$

If 2 circles cut each other orthogonally then

$d^2=r{1}^2+r{2}^2$ where d is the distance between the centers

$C_1=(1,11)$, $r1=\sqrt{1^2+{11}^2-5}=\sqrt{117}$

$C2=(-7,-3)$, $r2=\sqrt{7^2+3^2-k}=\sqrt{58-k}$

$(C1C2{)}^2=r{1}^2+r{2}^2$

$8^2+8^2=117+58-k$

$129=175-k$

$⟹k=47$