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An equilateral triangle ABC circumscribes the circle with equation $x^2 + y^2 = a^2$. The side BC of the triangle has equation $x = -a$.

a) Find the equations of AB and AC.

b) Find the equation of the circle circumscribing ABC.

Note: Pls show working when you answer and try to explain how you obtain the answer. By the way I am just a Year 10 student who is stuck on a textbook question and doing self-study before school starts.

My thoughts: I don't really know where to start to be honest. I tried to draw a diagram. $x = - a$ is a vertical line (parallel to y-axis) and is tangent to the given circle whose center is at the origin and radius is $a$. After drawing the diagram, I drew a triangle so that it circumscribes the circle and it looks like I need trigonometry to find AC and AB. Is that right? Since there is a right angle at the x-axis, do I use cos for AC and BC?

What I tried to do just now: Looking at the diagram, I attempted to use trigonometry for finding the equation of AC first. Since the inner circle's centre is on the origin, divided by the x-axis, I named M as the midpoint of BC. BM = CM has length $(\frac{- \alpha}{2})$. Also since the original equilateral triangle is now halved, it is a right angled triangle (there is a right angle and the other two angles has to be half the right angle). Now I use cos(45) $(\frac{- \alpha \div2}{h})$ but i kinda failed :( [h, for AB]

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  • $\begingroup$ Looks like someone asked the same question as me a few minutes ago hehe. Lol just realised $\endgroup$ – DragonReborn Jan 10 '16 at 4:15
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    $\begingroup$ Same comment applies: show us what you've done so far, so that we can better help you. No effort on your part probably means no effort on ours, either. $\endgroup$ – John Hughes Jan 10 '16 at 4:18
  • $\begingroup$ I'm stuck. I've explained what I did in the 'My thoughts' section. I need some guidance. I don't even know where to go next after drawing the diagram. $\endgroup$ – DragonReborn Jan 10 '16 at 4:20
  • $\begingroup$ The triangle is above x=-a. The top vertex, A is on y=? (x=0. Why?) Does this helps? $\endgroup$ – Moti Jan 10 '16 at 4:47
  • $\begingroup$ Why is that? Isn't x=-a a vertical line instead of horizontal? So shouldn't the triangle, with the top vertex, be pointing left lying on the x-axis (x,0)? The triangle sit on the base of the line tangent to the incircle. Hmm maybe I drew it wrong... $\endgroup$ – DragonReborn Jan 10 '16 at 5:03
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We can first determine the side length of $ABC.$ Since equilateral triangle $ABC$ circumscribes the circle, they must have the same center - the origin. Let point $P$ be the x-intercept of the line $x = -a.$ Then $OP = a.$ We know that $OP$ is a third of the median of triangle $ABC$ since it is equilateral. So the median length is just $3a.$ Using the 30-60-90 ratios, the side length of the triangle must be $\sqrt{3}a.$

a) Let $B$ be above $C.$ The coordinates for $B$ are $(-a, \frac{\sqrt{3}}{2}a).$ The coordinates for $A$ are $(2a, 0)$ because $OA$ is twice $OP.$ The equation of $AB$ is $\boxed{y = -\frac{\sqrt{3}}{6}x + \frac{\sqrt{3}}{3}}.$ By symmetry, the equation of line $CB$ is $\boxed{y = \frac{3}{6}x - \frac{\sqrt{3}}{3}}.$

b) Notice that $OA$ is a radius of the circumcircle of triangle $ABC$. We already found that $OA$ is $2a.$ So the equation of the circumcircle of triangle $ABC$ is $\boxed{x^{2} + y^{2} = 4a^{2}}.$ Notice that its center is also at the origin.

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  • $\begingroup$ Thx so much! It was very helpful :) $\endgroup$ – DragonReborn Jan 10 '16 at 6:16

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