Problema Solution
A triangle is inscribed in a circle. The vertices of the triangle are located at (-10, 10), (-6, 2) and (-2, 10). Using
your algebra skills (no constructions!), locate the center of the circle and then write an equation for the circle.
Answer provided by our tutors
let A(-10, 10), B(-6,2), C(-2,10)
First we will find the slopes of the lines AB and BC
AB: m1 = (2 - 10)/(-6 - (-10)) = - 8 / 4 = -2
BC: m2 = (10 - 2)/(-2 - (-6)) = 8/4 = 2
the center of the circle O lies in the intersection of the perpendicular bisectors of AB and BS
the equation of the perpendicular bisector of AB has slope that is equal to - 1/m1 and goes trough the middle of AB: C1((-10-6)/2, (10 + 2)/2) that is C1(-8, 6)
y - 6 = (-1/(-2))(x - (-8))
y = (1/2)x + 10
the equation of the perpendicular bisector of BC has slope that is equal to - 1/m2 and goes trough the middle of BC: A1((-6-2)/2, (2+10)/2) that is A1(-4, 6)
y - 6 = (-1/2)(x - (-4))
y = (-1/2)x + 4
we will find the coordinates of O by solving the system of equations
y = (1/2)x + 10
y = (-1/2)x + 4
x = -6
y = 7
click here to see the step by step solution of the equation
O(-6, 7) is the center of the circle
now we have to find the radius of the circle R. R is equal to the distance AO (R = AO)
A(-10, 10) and O(-6, 7)
R^2 = (-10 - (-6))^2 + (10 - 7)^2
R^2 = 16 + 9
R = 5
the equation of the circle is
(x - (-6))^2 + (y - 7)^2 = 5^2
(x + 6)^2 + (y - 7)^2 = 25