Coordinate Geometry: Are there enough information to find out the coordinates?

Question

Question: Given the circle $x^2+y^2=25$ is inscribed in triangle $\triangle ABC$, where vertex $B$ lies on the first quadrant. Slope of $AB$ is $\sqrt 3$ and has a positive y-coordinate, and $|AB|=|AC|$. Find the equations for $AC$ and $BC$

I found out the equation for the straight line passing through $AB$: Let the line be $y=\sqrt 3 x+c$. Then

$3x^2+2\sqrt 3 cx+c^2+x^2=25$

$\Delta =0$ (discriminant)

$(2\sqrt 3 c)^2 - 4(4)(c^2-25)=0$

$c=10$

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However, I don't see any simple way to find out the equations of line for $AC$ and $BC$. While it seems like there is enough information, I have tried using similar triangles, etc, but I can't find out the coordinates of the vertices. Can anyone give me some hints? Thank you!

Answer 1

Hint: Use the formula of incentre. It is given by $$(x,y)=(\dfrac{ax_1+bx_2+cx_3}{a+b+c},\dfrac{ay_1+by_2+cy_3}{a+b+c})$$

Here $a,b,c $ are the lengths of the sides of the triangle. Length a is opposite to the point A. Length b is opposite to the point B. Length c is opposite to the point C.
Here you can easily see $x=y=0$ & $b=c$

Answer 2

I will suggest to take $\alpha$ as parameter to write the equation. and you need to take care of the range of $\alpha$ .

Answer 3

I found it fairly straightforward to describe these lines with vector equations.

The line $AB$ is given as having a slope of $\sqrt3$, so it can immediately be parametrized as $\vec P_{AB}=\vec A+t\,(1,\sqrt3)$.

The line $AC$ is the reflection of this line in the line $AO$. A parametrization for this line can be found by reflecting the above parametrization of $AB$. Doing so yields $$\vec P_{AC}=\vec A+t\,\left[2{\vec A\cdot(1,\sqrt3)\over\vec A\cdot\vec A}\vec A-(1,\sqrt3)\right].$$

Finally, since the triangle is isosceles, the line $BC$ is perpendicular to $AO$ and tangent to the circle at the intersection of $AO$ with the circle. This suggests the equation $$\vec P_{BC}\cdot\vec A = \left(-5\frac{\vec A}{\|\vec A\|}\right)\cdot\vec A = -5\|\vec A\|.$$

There’s not enough information to determine a unique value for $A$. It must be farther from the $y$-axis than the point at which $AB$ intersects the circle, $\left(-\frac52\sqrt3,\frac52\right)$, because there’s only one tangent to the circle there and moving any closer takes $B$ out of the first quadrant. On the other hand, when $B$ is on the $y$-axis, the triangle is equilateral, putting $A$ at $(-5\sqrt3,-5)$, so $A$ must be closer to the $y$-axis than this. Combining these constraints gives $$A=\frac52[(-\sqrt3,1)-s\,(\sqrt3,3)]$$ for $0<s<1$. I’ll leave it to you to grind through the algebra and convert these equations into the required forms.

chenbai’s suggestion of using an angle as the parameter seems like a less messy way to go, though. We still derive the equations for $AC$ and $BC$ in the same way. With a slope of $\sqrt3$, $AB$ makes an angle of $\frac\pi3$ with the positive $x$-axis. $AC$ is the reflection of $AB$ with respect to $AO$, so we just need to find the angle of this line, which results in the equation $${y-y_A\over x-x_A}=\tan\left(2\alpha - \frac\pi3\right)$$ for $AC$. We need to express the coordinates of $A$ as a function of $\alpha$, but that’s just a matter of writing the equation for $AB$ in polar form. The bounds for $\alpha$ can be found by making the same considerations as in the vector-based solution above.

Similarly, $BC$ is orthogonal to $AO$, so its equation will be $${y+5\sin\alpha\over x+5\cos\alpha}=\tan\left(\alpha+\frac\pi2\right).$$ The plusses on the left side are because we want the opposite side of the circle from $A$.

Answer 4

$AB=AC$ suggests to me that line $BC$ is a vertical line and point $A$ is on the negative x-axis. See the picture.

$A$ has to be on the line perpendicular to segmanet $OP$ at $P$. Put $A$ in the second quadrant and $AB \lt AC$ Put $A$ in the third quadrant and $AB \gt AC$.

Let $P$ be the point where line $AB$ intersects the circle $x^2 + y^2 = 25$. If the slope of line $AB$ is $\sqrt 3$. Then the slope of line $OP$ must be $-\dfrac{1}{\sqrt 3}$. So the line $OP$ is the line $y = -\dfrac{1}{\sqrt 3}x$.

You said you just wanted a hint. So I will let you work out the rest.