SOLUTION: Find the equation of the circle whose center is on the line 2x - y + 4 = 0 and which passes thru the points (0,4) and (3,7) a. (x-7)^2 + (y-3)^2 = 8 b. (x-2)^2 + (y-3)^2 =

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Question 1189356: Find the equation of the circle whose center is on the line 2x - y + 4 = 0 and which passes thru the points (0,4)
and (3,7)
a. (x-7)^2 + (y-3)^2 = 8
b. (x-2)^2 + (y-3)^2 = 7
c. (x-7)^2 + (y-4)^2 = 6
d. (x-1)^2 + (y-6)^2 = 5
Found 3 solutions by math_tutor2020, Alan3354, ikleyn:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Answer: Choice D
(x-1)^2 + (y-6)^2 = 5

==================================================
How to get that answer:

Inspiration is drawn from this Numberphile video on YouTube
https://www.youtube.com/watch?v=wVH4MS6v23U
Even though the video topic is about triangle centers, it's still useful in determining the circle's center.
The particular section I have in mind is around 2:00 of that video. The professor talks about 2 towns and the goal of building a railroad that is perfectly smack dab in the middle of both towns. That way, each town has fair equal access to the tracks. In other words, you would travel the same distance to reach the railway from either town.

This railroad is the perpendicular bisector of segment AB, where A and B are the town locations.

In our case,
A = (0,4)
B = (3,7)

Let's find the perpendicular bisector of AB
First we'll need the slope of AB
m = slope
m = rise/run
m = (change in y)/(change in x)
m = (y2-y1)/(x2-x1)
m = (7-4)/(3-0)
m = 3/3
m = 1
The slope of AB is 1.

The slope of something perpendicular to this is -1.
Apply the negative reciprocal operation.
Rule: Any two perpendicular lines have their slopes multiply to -1, assuming neither line is horizontal nor vertical.

The perpendicular bisector has slope -1, but we don't know which point it goes through. To find one such point, we use the midpoint formula. This is because the "bisector" means we cut segment AB in half. Theres a point C such that AC = CB and C is on AB.

The x coordinate of point C is found by adding up the x coordinates of A and B, then dividing by 2.
x of C = (x of A + x of B)/2 = (0+3)/2 = 1.5
Same goes for the y coordinates
y of C = (y of A + y of B)/2 = (4+7)/2 = 5.5

The midpoint C is located at (1.5, 5.5)

So the equation of the perpendicular bisector of AB is
y-y1 = m(x-x1)
y-5.5 = -1(x-1.5)
y-5.5 = -x+1.5
y = -x+1.5+5.5
y = -x+7
Conveniently, the decimal portions add up to a whole result.

If town A is at (0,4) and town B is at (3,7), then we can build a perfectly fair railroad track along the line y = -x+7 so that people from either town are the same distance away from the track.

Put another way: Let T represent the location of the train.
It can be anywhere on the track (not necessarily at the same location as point C).
The distance from T to either town is the same.
segment AT = segment BT

Now you may or may not have noticed, but the given equation
2x-y+4 = 0
is also another railroad track equation. However, we don't have a third city to deal with.
For more information, search out "circumcenter" and "circumcircle".

But intersecting these two railroad track equations will pinpoint the center of the circle.

Plug y = -x+7 into the equation your teacher gave you
2x-y+4 = 0
2x-(y)+4 = 0
2x-(-x+7)+4 = 0
2x+x-7+4 = 0
2x+x-3 = 0
3x-3 = 0
3x = 3
x = 3/3
x = 1

This leads to,
y = -x+7
y = -1+7
y = 6
The center of the circle is (1,6)
So (h,k) = (1,6) i.e. h = 1 and k = 6

The template
(x-h)^2 + (y-k)^2 = r^2
represents any circle with center (h,k) and radius r
Plug in those h & k values
(x-h)^2 + (y-k)^2 = r^2
(x-1)^2 + (y-6)^2 = r^2

We only have one answer choice that matches (choice D), so we technically could be done at this point.
However, let's assume we didn't have those convenient multiple choice answers in front of us.

The question is now: How can we find the radius?
We do this by computing the distance from the center (1,6) to either A or B.
If the center (1,6) is point D, then DA = DB as they are radii of the same circle.

Apply the distance formula for (1,6) and (0,4)
d = sqrt( (x1-x2)^2 + (y1-y2)^2 )
d = sqrt( (1-0)^2 + (6-4)^2 )
d = sqrt( 1^2 + 2^2 )
d = sqrt( 1 + 4 )
d = sqrt( 5 )
The radius is exactly r = sqrt(5) units long
Squaring both sides gets us r^2 = 5
You should get the same r^2 value if you were to find the distance from (1,6) to (3,7).

So to wrap everything up, we have
(x-1)^2 + (y-6)^2 = r^2
turn into
(x-1)^2 + (y-6)^2 = 5

This confirms the answer is definitely choice D.

Graph:

The graph was made with GeoGebra.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find the equation of the circle whose center is on the line 2x - y + 4 = 0 and which passes thru the points A(0,4) and B(3,7)
------------------------
The slope of AB is +1.
The center has to be on the perpendicular bisector of the line thru the 2 points.
The midpoint of AB is (1.5,5.5) and it's slope is -1.
y-5.5 = -1*(x-1.5) = -x + 1.5
y = -x + 7
=====================
The center is the intersection of the 2 lines.
2x - y + 4 = 0
y = 2x+4
y = -x + 7
---
2x+4 = -x+7
3x = 3
x = 1, y = 6 ----> center is (1,6)
r = the distance from the center to either point.
==========================================
Choice d. r = 5
------------------------------------------
There's no good reason for having 4 (or any #) of choices.
The problem should be, "Find the equation of the circle."
Answer by ikleyn(52754)   (Show Source): You can put this solution on YOUR website!
.
Find the equation of the circle whose center is on the line 2x - y + 4 = 0 and which passes
thru the points (0,4) and (3,7)
a. (x-7)^2 + (y-3)^2 = 8
b. (x-2)^2 + (y-3)^2 = 7
c. (x-7)^2 + (y-4)^2 = 6
d. (x-1)^2 + (y-6)^2 = 5
~~~~~~~~~~~~~


            I read,  interpret,  understand and treat this problem differently from two other preceding tutors.

            I read this problem in a way as if it asks me to choose between the four given alternatives.
            I see that I can quickly identify one of the four alternatives simply by substituting coordinates of the point (0,4)
            into the given equations. 


(a)  (0-7)^2 = 7^2 = 49 (too much) just tells me that this equation does not work.



(c)  The same is with the equation (c).



(b)  (0-2)^2 + (4-3)^2 = 2^2 + 1^2 = 5 tells me that equation (b) does not work.



(d)  (0-1)^2 + (4-6)^2 = 1^2 + 2^2 = 5 tells me that equation (d) is, probably, the only good candidate.


     To get the final decision, I only shoud check that the second points' coordinates (3,7) satisfy that equation.


     (3-1)^2 + (7-6)2 = 2^2 + 1^2 = 5  confirm that both given points lie on the circle.



(*)  As the last check, I should make sure that the center of this circle, which is (1,6) (from equation (d)) 

     lies at the given line; so I substitute the coordinates of the center (1,6) into the equation

     of the line , and I see that 2*1 - 6 + 4 = 0, so this statement is valid.


At this point,  all checks are done and the problem is just solved  (practically,  mentally,  in the head).

----------------

In the given form,  the problem is the kind of  " joke Math problems "  in this area - - - for those
who does understand Math jokes.



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