Find center of circle, which tangent to $y=2$ at $(3,2)$ and $y=-x\sqrt 3 +2$

Question

Using $$r=\frac{|Ax + By + c|}{\sqrt{A^2+B^2}}$$

I get $$b=2+a\sqrt3$$ and stuck,because the options are on numbers. What do i do next?

Answer 1

Suppose $C=(a,b)$ is the center of the circle.

Then $a=3$ because the circle is tangent to $y=3$ at point $(3,2)$.

Also the distance of $C$ from both lines $y=2$ and $y=-x\sqrt{3}+2$ is same: $$|b-2|=\frac{|b+3\sqrt{3}-2|}{2}$$ and note that we have two solutions for $b<2$ and $b>2$.

can you proceed?

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Edit: If $b>2$, then $b+3\sqrt{3}-2>3\sqrt{3}>0$, thus we have $$b-2=\frac{b+3\sqrt{3}-2}{2}$$ which gives $b=2+3\sqrt{3}$.

If $b<2$, $|b-2|=2-b$, but $|b+3\sqrt{3}-2|$ is $\pm(b+3\sqrt{3}-2)$

thus $$2-b=\frac{b+3\sqrt{3}-2}{2}$$ which gives $b=2-\sqrt{3}$ or $$2-b=-\frac{b+3\sqrt{3}-2}{2}$$ which leads to repeated value $b=2+3\sqrt{3}$

Answer 2

Tangent lines meet at $B=(0,2)$, and the minor angle at $B$ is $60^\circ$, so we can construct an equilateral $\triangle ABC$, for which the minor and major circles tangent to $y=2$ at $T=(3,2)$ are the inscribed and escribed circles, respectively.

The side length of the equilateral $\triangle ABC$ $|BC|=|CA|=|AB|=a=b=c=|AB|=2\cdot|BT|=6$.

So, the radius of the inscribed circle is \begin{align} r&=\tfrac13\cdot a\cdot\tfrac{\sqrt3}2 =\sqrt3 . \end{align}

And the radius of the escribed circle is \begin{align} r_c& =\frac{a+b+c}{a+b-c}\cdot r =\frac{3\,a}{a}\cdot r =3\,r , \end{align}

so the coordinates of the centers are

\begin{align} I&=(3,2-\sqrt3) \approx(3,0.268) ,\\ O_c&=(3,2+3\sqrt3) \approx(3,7.196) . \end{align}

Answer 3

You can think of the entire thing geometrically. If you had two lines that were tangent to a circle and you wanted to draw said circle, here would be the steps:

1. Construct a perpendicular line from a point on one of the given lines.
2. Construct an angle bisector of the given lines.
3. The constructed lines meet at the centre of the circle.

So this case is no exception. The thing is, you have to be smart about it.

Consider $f(x)=1 \cdot x$. This line forms an angle of $45^\circ$ with any horizontal line you draw on the plane(why?) But $\tan{45^\circ}=1$! This means that the slope of a line is intrinsically connected with the tangent of the angle it forms with a horizontal. No, it is the tangent of that angle!

The line $y=x(-\sqrt{3})$ forms an angle of $-60^\circ$ or $120^\circ$ with the horizontal and thus with the other given line. Therefore so should it's parallel counterpart $y=x(-\sqrt{3})+2$ which we ignored for a while. Thus the angle bisector line has a slope of $\tan{-30^\circ}$. It should also pass through the common point $(0,2)$ so it's y-intercept is 2.

Equation of line 1 $\implies y=-x\dfrac{\sqrt{3}}{3}+2$

The second one is simple. From the point $(3,2)$, the perpendicular line would be:

Line 2 $\implies x=3$

Where the two lines meet is where the x value on line 1 $=3$, so the y value $=\dfrac{-3\sqrt{3}}{3}+2=2-\sqrt{3}$

What's the centre of the circle?

$(3, 2-\sqrt{3})$

What's the radius?

$=2-(2-\sqrt{3})=\sqrt{3}$

Have fun with those.

Edit: I just realised you could have drawn another circle on the other side of the lines. But it's the co-ordinate given that determines which circle we want. But I guess for one with the same radius use $(-3,2)$ instead