Hi, have a few review problems for a test tomorrow, want to make sure I can understand everything.

1. Given the points below, show using coordinate geometry that BC and EF represent the diameters of the circles that contain the points ABC and DEF.
A (-13, -12), B(-10, -3), C(-7, -14)
D (6, 2), E(-2, 0), F(9, -10)

Find the equation of the circles that contain the points ABC and DEF.

I have this problem and I'm not 100% sure if I'm doing it right. My plan was to find the slopes of AB, BC, AC, DE, DF, and EF. 4 of them would be perpendicular, and I would use this to assume that because this is a right angle, the third side forms a right triangle and must be the diameter.

I'm guessing that there are two circles, one with ABC, and the other with DEF? Would I just first prove that BC/EF are the diameters, find the midpoint/distance afterwards? Or is there something I'm missing, or incorrect proof?

2. Find the value(s) of k that make the graphs of the given equations intersect in exactly one point.
y = x^2 + 4x - 1
y = 3x + k

Would I just set them equal to each other and use the quadratic formula?

Thanks!

I would use the coordinates to get the equation of the circle, then you can show geometrically that a straight line exists from B to C that passes through the center defined by the equation of the circle (diameter definition).

Set the two equations equal to one another. For it to intersect at exactly one point (tangent), there can be only one X value where they are equal. This should lead you to that specific k value that makes this true.

This post was edited by Dontrunaway on Nov 25 2014 07:51pm

Alright cool thanks, is using the slope method sufficient? Or are there any other methods that can be used?

You have to show geometrically that the center of the circle you found is equidistant to all points given, then just showing that 2 points + center lie on the same line (y = mx + should be sufficient to define that line as the diameter.

Question 1

If you find the intersections of the perpendicular bisectors of AB and BC you will find the point of the center of the circle
The line AB has midpoint (-13-10,-12-3)/2
= (-23/2,-15/2)
Slope of line AB is [-3 – (-12)] / [-10 – (-13)]
= 9/3
= 3

The slope of the perpendicular bisector AB is
-1/3

Now we can find the equation of the line in the form y = mx + b
-15/2 = -1/3(-23/2) + b
b = -34/3

therefore the equation of the line is
y = -(1/3)x – (34/3).

Follow the same steps to find the equation of the perpendicular bisector of BC
You will get
y = (3/11)x – (68/11)

equating the 2 equations to solve for x and y, you get
x = -17/2 y = -17/2 which are your coordinates for the center of the circle

The equation of the circle is therefore
(x+(17/2))^2 + (y+(17/2))^2 = r^2

To solve for the radius, sub in points you know are on the circle. Ie (-10, -3)
(-10+(17/2))^2 + (-3+(17/2))^2 = r^2
r = √ (65/2)

To show BC is the diameter, find the distance of BC
|BC| = √[(-7 – (-10))^2 + (-14 – (-3))^2]
= √130

|BC| is twice the radius and are points on the circle, therefore it must be the diameter. The same steps can be applied for the other circle.
Note that this is just one of the MANY ways to solve this question.

Question 2

Equate the equations
x^2 + 4x – 1 = 3x + k

rearrange
x^2 + x – 1 – k = 0

There is only one solution to this polynomial iff the discriminant is 0
Δ = b^2 – 4ac = 0
1 – 4(-1-k) = 0
1 + 4 + 4k = 0
k = -5/4

This post was edited by cdexswzaq on Nov 29 2014 11:46pm