Question

For the given two circles $x^2+y^2=1$ and $x^2+y^2-2x-6y+6=0$ the line $4x-3y-5=0$ is

• A. A direct common tangent to the cirlces
• B. An indirect common tangent to the circles
• C. A tangent to the first circle only
• D. A tangent to the second cricle only

Answer 1

The correct option is A A direct common tangent to the cirlces

For the circle $\underset{c_1}{\underset{\downarrow }{x^2+y^2=1}}$ and $\underset{c_2}{\underset{\downarrow }{x^2+y^2-2x-6y+6=0}}$ the line

$4x-3y-5=0$

so, circle (i) $x^2+y^2=1$ center $(0,0)$, radius $=1$

now to be a tangent the perpendicular from centre would be equal to radius.

so, perpendicular from $(0,0)$ on $(4x-3y-5)$

$=\left|\frac{4\left(0\right)-3\left(0\right)-5}{\sqrt{4^2+3^2}}\right|=\left|\frac{-5}{5}\right|=1$.

so it is also tangent of $c_2$

Now check that it is direct or indirect so (Ref. image)

so, $(0,0),(3,3)$ $\Rightarrow (y-0)=\frac{3}{1}|x-0|$ $y=3x$

and $4x-3y-5=0$

$y=3x$ .... intersection point $x=-1$ $y=-3$

If it lies between $(0,0)$ and $(1,3)$ then indirect.

So it is direct common tangent