Question

a. Find the center of the circle. b. Find the radius. If your answer is not integer, express it in radical form. c. Write an equation for the circle.

Answer 1

I apologize here is the whole thing.

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

a. Find the center of the circle.
b. Find the radius. If your answer is not integer, express it in radical form.
c. Write an equation for the circle.

Answer 2

mid point is center.

Answer 3

I am not very good at Geometry so I struggle with this, and my book isn't very much help when I go to look in it i end up with more questions than answers.

Answer 4

find the mid point.
\[x=\frac{ x _{1} +x _{2}}{ 2 }\]
\[y=\frac{ y _{1}+y _{2} }{ 2 }\]

Answer 5

That is the formula for midpoint?

Answer 6

yes

Answer 7

O.K. so then it would look like this

Answer 8

-10 + 4 over 2

Answer 9

and -2 + 4 over 2

Answer 10

Right?

Answer 11

\[y=\frac{ -2+6 }{ 2 }=?\]

Answer 12

for x ,it is correct.

Answer 13

oops that is what I meant.

Answer 14

now what is the center.

Answer 15

so the center is (-3,2)

Answer 16

correct
find the distance between the center and one point of diameter.

Answer 17

do you know the formula for distance?

Answer 18

So I use the distance formula and input (-3,2) and either (-10,-2) or (4,6)

Answer 19

Correct?

Answer 20

yes

Answer 21

O.K. and just to make sure I remember the distance formula is

Answer 22

\[\sqrt{( x _{1}- x _{2})^{2} + (y _{1}-y _{2})^{2}}\]

Answer 23

Right?

Answer 24

yes

Answer 25

correct.

Answer 26

So the answer would be 8.06?

Answer 27

Is 8.06 my radius?

Answer 28

keep it in square root form

Answer 29

yes, but sqr(65) is more accurate

Answer 30

\[\sqrt{65}\]

Answer 31

Oh O.K I see

Answer 32

now write the eq. of circle

Answer 33

So then with making the equation I need a little extra guidance.

Answer 34

the general form for the equation of a circle is
\[ (x-h)^2 + (y-k)^2 = r^2\]
where (h,k) is the center and r is the radius

Answer 35

\[\left( x-h \right)^2+\left( y-k \right)^2=r^2\]

Answer 36

So mine would be (x-(-3))^2 + (y-2)^2= √65 ^2

Answer 37

correct.
simplify it

Answer 38

Simplify what the exponents?

Answer 39

the square of a square root undoes the square root.

Answer 40

Right I knew that.

Answer 41

also, most people would simplify x- -3 to x+3

Answer 42

Oh!

Answer 43

Thank you very much to you both I wish I could give two Best Responses.

Answer 44

:^)

Answer 45

you can also write the eq. of circle from the extremities of diameter by the formula
\[\left( x-x _{1} \right)\left( x-x _{2} \right)+\left( y-y _{1} \right)\left( y-y _{2} \right)=0\]

Answer 46

y does it equal 0?

Answer 47

yes

Answer 48

Oh alright I will write that down. Thanks again!! X^D

Answer 49

eq. of circle in general form is
\[x^2+y^2+2gx+2fy+c=0\]
where center is (-g,-f)
and \[radius=\sqrt{g^2+f^2-c}\]