Question

If the line $ax+by=2,(a\ne 0)$ touches the circle $x^2+y^2-2x=3$ and is normal to the circle $x^2+y^2-4y=6$, then $a+b$ is equal to

• A. $\frac{-4}{3}$
• B. $\frac{-5}{3}$
• C. $\frac{-1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{-1}{3}$

Consider the given equation of the circle.

$\begin{array}{c}{x}^2+y^2-2x=3\\ {\left(x-1\right)}^2+y^2=4\\ Centre\left(1,0\right)\\ Radius=2\\ and,\\ ax+by=2\\ x^2+y^2-4y-6=0\\ Centre=\left(0,2\right)\\ 2b=2\\ b=1\\ Now,\\ \frac{\left|a-2\right|}{\sqrt{a^2+b^2}}=2\\ {\left(a-2\right)}^2=4\left(a^2+1\right)\\ a^2+4-4a=4a^2+4\\ 3a^2+4a=0\\ a\left(3a+4\right)=0\\ a\ne 0\\ a=-\frac{4}{3}\\ a+b=-\frac{1}{3}\\ Hence,optionCisthecorrectanswer.\end{array}$