Question

The condition that the straight line $cx-by+b^2=0$ may touch the circle $x^2+y^2=ax+by$, is:

• A. $abc=1$
• B. $a=c$
• C. $b=ac$
• D. None of these

Answer 1

The correct option is B $a=c$

Straight line,

$cx-by+b^2=0$

Slope of straight line,

$\frac{dy}{dx}=\frac{c}{b}⟶\left(1\right)$

We have circle as,

$x^2+y^2=ax+by⟶\left(2\right)$

Diff. curve $\left(2\right)$ w.r.t. $x$,

$2x+2y\frac{dy}{dx}=a+b\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{2x-a}{b-2y}⟶\left(3\right)$

From circle, for $r=0\Rightarrow y=b$

Put in $\left(3\right)$ and equate to $\left(1\right)$

We get,

$\frac{-a}{b-2b}=\frac{c}{b}$

$\Rightarrow a=c$