1. ## equation of lines

Consider the points A(6,-1), B(3,2), C(0,-1).
(a) Find the equation of the circle S passing through A,B and C (b) Determine the center and radius of S
(c) Find a fourth point D on S such that ABCD is a square
(d) Write down the equation of the tangent line to S at D.

2. Find the point of intersection for the line that is the perpendicular bisector of the line segment AB and the perpendicular bisector of the line segment BC. The point is the center of the circle. The distance from A to that point is the radius of the circle.

3. Originally Posted by watcher
Consider the points A(6,-1), B(3,2), C(0,-1).
(a) Find the equation of the circle S passing through A,B and C (b) Determine the center and radius of S
(c) Find a fourth point D on S such that ABCD is a square
(d) Write down the equation of the tangent line to S at D.
You can use (c) to "cheat" in (a) and (b)! If (c) is possible, that is, if there exist a point D such that ABCD forms a square, then the two line segments, AB and BC, must be perpendicular and of the same length.

So check that- The distance from A(6,-1) to B(3,2) is $\sqrt{(6-3)^2+ (2+1)^2}= \sqrt{9+ 9}= 3\sqrt{2}$ and the slope of the line is $(2+1)/(3-6)= -1$. The distance from B(3,2) to C(0,-1) is $\sqrt{(3-0)^2+ (2+1)}= \sqrt{9+ 9}= 3\sqrt{2}$ and the slope of the line is $(2+1)/(3-0)= 1$. Yes, A, B, and C do form 3 vertices of a square. But that means that AC is a diagonal of that square and so a diameter of the circle through A, B, and C. The center of the circle is the midpoint of ((6+0)/2, (-1-1)/2)= (3, -1) and the radius is the distance from that point of any of A, B, and C which is also half the distance from A to C, $\sqrt{(6-0)^2+ (-1+ 1)^2}/2= 3$