# Thread: Circles and Angle Bisectors again help plsss...

1. ## Circles and Angle Bisectors again help plsss...

Hi, im still confused about inscribed circles , well heres the prob:

The sides of a triangle are on the lines $6x-7y+11=0$, $2x+9y+11=0$, and $9x - 2y - 11 = 0$. Find the equation of the circle inscribed in the triangle...
And could you use what soroban did in this thread?
http://www.mathhelpforum.com/math-he...bisectors.html

This's different from what he did When using the equation of directed distance here the square root of the denominatora is different and i dont know how to equate it ...

Any help be appreciated

Hi, im still confused about inscribed circles , well heres the prob:

The sides of a triangle are on the lines $6x-7y+11=0$, $2x+9y+11=0$, and $9x - 2y - 11 = 0$. Find the equation of the circle inscribed in the triangle...
And could you use what soroban did in this thread?
http://www.mathhelpforum.com/math-he...bisectors.html

This's different from what he did When using the equation of directed distance here the square root of the denominatora is different and i dont know how to equate it ...

Any help be appreciated
One thing you could do, since you have the equations for the lines is to
1) Find the three points of intersection representing the points of the triangle
2) Find the equation of the (unique) circle that contains those points.

I'm kind of in a rush right now, but if you need more help simply let me know.

-Dan

Hi, im still confused about inscribed circles , well heres the prob:

The sides of a triangle are on the lines $6x-7y+11=0$, $2x+9y+11=0$, and $9x - 2y - 11 = 0$. Find the equation of the circle inscribed in the triangle...
...
Hello,

the centre of the inscribed circle is $C(c_1, c_2)$. It has the same distance to all three straight lines. The formula to calculate the distance of a point P(p_1, p_2) to a line is:

$d=\frac{Ap_1+Bp_2+C}{\sqrt{A^2+B^2}}$

You get 3 equations:

$\frac{6m_1-7m_2+11}{\sqrt{36+49}}=r$
$\frac{2m_1+9m_2+11}{\sqrt{4+81}}=r$
$\frac{9m_1-2m_2-11}{\sqrt{81+4}}=r$

Luckily the denominator is the same in all three equations. Now solve for m_1, m_2 and r. (It's not so easy!). I've got the result x = 0, y = 0, r = 11/(√85). Therefore the equation of the inscribed circle is:

$x^2+y^2=\frac{121}{85}$

I've attached a diagram of the situation.

EB

No, I haven't solved it yet.
I'm sure that distance-to-a-line approach will still work.

I need to make an accurate sketch
. . to get an idea where the incenter is.

I "eyeballed" an interesting fact.
Look at the coefficients of the second and third equations.
. . $\begin{array}{ccc}L_1 \\ L_2 \\ L_3\end{array}
\begin{array}{ccc}6x-7y+11 & = & 0\\2x+9y+11 & = & 0 \\ 9x - 2y - 11 &= &0\end{array}$
. . $L_2$ and $L_3$ are perpendicular.

Will that help? .I have no idea . . .

Sheesh!. . EB already nailed it!

5. Originally Posted by Soroban

No, I haven't solved it yet.
I'm sure that distance-to-a-line approach will still work.

I need to make an accurate sketch
. . to get an idea where the incenter is.

I "eyeballed" an interesting fact.
Look at the coefficients of the second and third equations.
. . $\begin{array}{ccc}L_1 \\ L_2 \\ L_3\end{array}
\begin{array}{ccc}6x-7y+11 & = & 0\\2x+9y+11 & = & 0 \\ 9x - 2y - 11 &= &0\end{array}$
. . $L_2$ and $L_3$ are perpendicular.

Will that help? .I have no idea . . .

Sheesh!. . EB already nailed it!
Of course the distance-to-a-line approch still works here.
I did it a while ago but I was getting the cennter of a circle that is outside of the enclosed triangle. Its center was at (88/17, 22/17).
So I looked at my sketch--it is not accurate drawing--and guessed what if (0,0) were the center. The center in my sketch is very close to (0,0).
So I played with:
r1 = |6*0 -7*0 +11| / sqrt(6^2 +(-7)^2) = 11/sqrt(85)

r2 = |2*0 +9*0 +11| / sqrt(2^2 +9^2) = 11/sqrt(85)

r3 = |9*0 -2*0 +11| / sqrt(9^2 +(-2)^2) = 11/sqrt(85)

So the 3 radii came out equal if the center were at (0,0).

Therefore, the inscribed circle is x^2 +y^2 = 121/85. -------answer.