The area of equilateral triangle inscribed in the circle x^2+y^2-4x+8y+11=0 is of the form a*sqrt(b)/c. Find a+b+c.
To find the circumcenter, we first need to know the radius of the circle.
We can rewrite the circle equation to: x2−4x+y2+8y=−11
We can now complete the square, and we get: (x−2)2+(y+4)2=4+16−11
The left hand is equal to 9, meaning the radius is 3.
We now know the circumcenter is 3.
The equation for the circumcenter is: abc4[ABC]
But, we have an equilateral triangle, so a=b=c. This means we can substitute both b and c for a.
This gives us the equation: a34[ABC]=3
The formula for the area of an equilateral triangle is √34s2
We can substitute this in, and we get: a34√34a2=3
We can simplify and divide by a2 to get: a√3=3
This means that the side of the triangle is 3√3.
Can you find the area from here?
Hint: Use the formula for the area of the triangle