Question

If the straight line $ax+by=2$; a,b $\ne 0$ touches the circle $x^2+y^2-2x=3$ and is normal to the circle $x^2+y^2-4y=6$, then the values of $a$ and $b$ are

• A. $a=1,b=2$
• B. $a=1,b=-1$
• C. $a=\frac{-4}{3},b=1$
• D. $a=2,b=1$

Answer 1

Answer: (C)

$a=\frac{-4}{3},b=1$

Given $ax+by=2$ is tangent to $x^2+y^2-2x-3=0$
so, $r=\sqrt{4}=|\frac{a-2}{\sqrt{a^2+b^2}}|$
$4(a^2+b^2)=(a-2{)}^2$ ---{1}
and , $ax+by=2$ is also normal to $x^2+y^2-4y-6=0$
so, $ax+by=2$ passes through (0,2) center of circle.
$b\times 2=2$
$b=1$
so from (1) , $4a^2+4=(a-2{)}^2$
$4a^2+4=a^2-4a+4$
$3a^2=-4a$
$a=\frac{-4}{3}$ $[Givena,b\ne 0]$
$so,a=-\frac{4}{3}$ and $b=1$