anonymous
  • anonymous
A diameter of a circle has endpoints at (4, 6) and the origin. Which point is also on the circle? A. (–1, 2) B. (6, 3) C. (5, 0) D. (–1, 1) E. NOTA Please explain.
Mathematics
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anonymous
  • anonymous
A diameter of a circle has endpoints at (4, 6) and the origin. Which point is also on the circle? A. (–1, 2) B. (6, 3) C. (5, 0) D. (–1, 1) E. NOTA Please explain.
Mathematics
katieb
  • katieb
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campbell_st
  • campbell_st
find the midpoint x = (4 + 0)/2 y = (6 + 0)/2 that will give the centre
campbell_st
  • campbell_st
then the equation is (x - h)^2 + (y-k)^2 = r^2 in this case r^2 = 52 (h, k) is the centre... sub in each point... to test
anonymous
  • anonymous
answer is D (-1,1) please can you show more steps

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anonymous
  • anonymous
I didn't get it
Directrix
  • Directrix
|dw:1328772724764:dw|
anonymous
  • anonymous
Please explain and show me the steps
anonymous
  • anonymous
lol beauty picture
Directrix
  • Directrix
@Sammy - I lost power but have returned to explain the problem.
anonymous
  • anonymous
after getting midpoint what to do
anonymous
  • anonymous
Thank you so much Directrix.
Directrix
  • Directrix
The midpoint of the diameter is the center of the circle. One-half the length of the diameter is the radius of the circle.. We need that for the general equation of the circle: (x - h)^2 + (y-k)^2 = r^2 and then ...
Directrix
  • Directrix
From (0,0) to (2,3) is the length we need. The distance formula gives r = Square root[ ( 2 - 0)^2 + 3 - 0)^2] r = Square Root [ 4 + 9] = Square root of 13. And, then ...
Directrix
  • Directrix
r = √ 13 and (h,k) = (2,3)
Directrix
  • Directrix
(x - h)^2 + (y-k)^2 = r^2 (x - 2) ^2 + (y - 3) ^ 2 = (√ 13) ^2 (x - 2) ^2 + (y - 3) ^ 2 Option checking time now ...
Directrix
  • Directrix
(x - 2) ^2 + (y - 3) ^ 2 I am testing the D option D. (–1, 1) In the circle equation, substitute -1 for x and 1 for y and see if the left side = the right side of 13.
Directrix
  • Directrix
(x - 2) ^2 + (y - 3) ^ 2 ( (-1- 2)^2 + (1 - 3) ^2 = (-3) ^2 + (-2) ^2 = 9 + 4 = 13. So, the point (-1,1) is on the circle.
Directrix
  • Directrix
Suppose we had tested option A. (–1, 2). Then, ( (-1 -2) ^2 + ( 2 - 3) ^ 2 = (-3)^2 + (-1) ^2 = 9 + 1 = 10 which is not equal to 13. So, the point (-1,2) is NOT on the circle.
Directrix
  • Directrix
Questions?
anonymous
  • anonymous
yes.
Directrix
  • Directrix
OK
anonymous
  • anonymous
( (-1- 2)^2 + (1 - 3) ^2 = (-3) ^2 + (-2) ^2 = 9 + 4 = 13. but it's not sqrt13
Directrix
  • Directrix
Remember the formula (x - 2) ^2 + (y - 3) ^ 2 = (√ 13) ^2 (x - 2) ^2 + (y - 3) ^ 2 = (√ 13) ^2) The (√ 13) ^2 = 13. We are testing against 13.
anonymous
  • anonymous
Thank you so much for explaining and your time
Directrix
  • Directrix
Any time. Glad to help.
anonymous
  • anonymous
Can I ask one more problem?
Directrix
  • Directrix
Okay
Directrix
  • Directrix
Is it already posted?
anonymous
  • anonymous
Twice an integer increased by fourteen is less than 5. What is the greatest solution? Please explain. it's a multiple choice A. -5 B.-4 c.-4.5 D.- 4.6 E.NOTA
anonymous
  • anonymous
I already posted.but I didn't get the correct answer
anonymous
  • anonymous
Answer is A
Directrix
  • Directrix
Let x = the integer Then, 2x + 14 < 5
Directrix
  • Directrix
2x + 14 < 5 2x < 5 - 14 2x < -9
Directrix
  • Directrix
The answer is an integer. Remember that. 2x < -9 x < - 4.5 Option time.
Directrix
  • Directrix
What is the biggest integer LESS than - 4.5?
Directrix
  • Directrix
|dw:1328774992369:dw|
anonymous
  • anonymous
so it is -5
Directrix
  • Directrix
The biggest integer less than -4.5 is -5.
anonymous
  • anonymous
Thank you so much. I didn't think like that.
Directrix
  • Directrix
If the question had ended up as x >- 4.5, we would select the first integer to the right of -4.5 which would be -4. On a number line, always remember that the greater of two numbers lies to the right of the other.
Directrix
  • Directrix
"I didn't think like that." The simplest things can be the most confusing.
anonymous
  • anonymous
That' right!
anonymous
  • anonymous
Thank you so much for your help!
Directrix
  • Directrix
You are welcome. Any time.

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