Question

If the circles $x^2+y^2+2a^{1}x+2b^{1}y+c^1=0$ and
$2x^2+2y^2+2ax+2by+c=0$ intersect orthogonally, then

• A. $a{a}^1+b{b}^1=c+c^1$
• B. $a{a}^1+b{b}^1=c+\frac{c^1}{2}$
• C. $a{a}^1+b{b}^1=\frac{c}{2}+c^1$
• D. $2(a{a}^1+b{b}^1)=c+c^1$

Answer 1

Answer: (C)

$a{a}^1+b{b}^1=\frac{c}{2}+c^1$

Given equations of circles in standard form are: (make coefficients of square terms 1.)
$x^2+y^2+2a^{1}x+2b^{1}y+c^1=0$ and
$x^2+y^2+ax+by+\frac{c}{2}=0$
For orthogonality
$2(a^1\frac{a}{2}+b^1\frac{b}{2})=c^1+\frac{c}{2}$
$\Rightarrow a^{1}a+b^{1}b=c^1+\frac{c}{2}$