Thread: equation

hi i've got the question:

find the equation of the circle which passes through the points A(2, -1) B(4, -1) and C(0, 1)

so i started by finding the mid points of two of the lines so i could use the perpendicular bisectors to find the centre. the midpoint of AB is (3, -1) and the midpoint of AC is (1, 0)

then i tried finding the gradients of AB and AC. and came up with gradient = 0 for AB. and came up with -1 for AC. how would you express this as a gradient and how would you express the perpendicular? and how would you express the equations for both

basically i'm asking, can someone show me step by step how to complete the question

thanks

I'm afraid I don't understand what you're asking...? You say you found the gradients ("slopes", for American readers), but then you ask how to express the gradients...?

You have three points on the circle, and you have the general circle equation:



You could try plugging the three x,y-points into this equation, and then solving the resulting non-linear system for h, k, and r. But the method you've suggested may be easier.

Since points A and B lie on the horizontal line y = -1, their perpendicular bisector will be of the form x = (some number). So use the midpoint between A and B, and take the x-coordinate for your line equation.

You have found the midpoint of AC. To learn the relationship between the slopes of lines and their perpendiculars, try here. To learn how to find the line equation for the perpendicular bisector of AC from the slope and the midpoint, try here.

You should end up with the intersection of x = 3 and y = -x + 1 being the center (h, k) of your circle. Then the distance between the center and any of the three given points will give you the length r of the radius.