Question

lf $(1,2)$ and $(3,4)$ are limiting points of the given coaxial system then the least circle belonging to the orthogonal coaxial system is $x^2+y^2+ax+by+c=0$. Then $(a,c)=$

• A. $(-4,11)$
• B. $(-6,11)$
• C. $(4,11)$
• D. $(4,-11)$

Answer 1

The correct option is A $(-4,11)$

$(1,2)$ and $(3,4)$ are limiting point of coaxial system
for least circle belonging to this system has centre
mid-point of limiting point.
So, $(2,3)$ is a centre of circle
$(x-2{)}^2+(y-3{)}^2=(\sqrt{1^2+1^2}{)}^2$
$x^2+y^2-4x-6y+13=2$
$\Rightarrow x^2+y^2-4x-6y+11=0$
So, $a=-4$ and $c=11$
$\Rightarrow (a,c)=(-4,11)$