# Thread: Finding equation of circle.

1. ## Finding equation of circle.

Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

2. Originally Posted by behindDkeys
Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

The mid point is the center of the circle

$(x-x_0)^2+(y-y_0)^2=r^2$

3. Originally Posted by behindDkeys
Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius ( $r$).

Find the midpoint of the segment between these two points. This is your centre $(h, k)$.

The formula of the circle is

$(x - h)^2 + (y - k)^2 = r^2$.

4. Originally Posted by Prove It
Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius ( $r$).

Find the midpoint of the segment between these two points. This is your centre $(h, k)$.

The formula of the circle is

$(x - h)^2 + (y - k)^2 = r^2$.

ok so what i do is i take the points and put it on an A^2+B^2=C^2?

coz i dont really get pythagoras on this i dont see ABC

5. Once you have the mid point, you can figure out the distance to one of the points using the distance formula. The distance you will have solved for will be the radius.

6. Originally Posted by behindDkeys
ok so what i do is i take the points and put it on an A^2+B^2=C^2?

coz i dont really get pythagoras on this i dont see ABC
The distance formula

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

is a direct application of Pythagoras' Theorem.

7. Originally Posted by dwsmith
Once you have the mid point, you can figure out the distance to one of the points using the distance formula. The distance you will have solved for will be the radius.
uh can you show me step by step
like plug in the given then ima solve the other stuff

8. Originally Posted by Prove It
The distance formula

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

is a direct application of Pythagoras' Theorem.

oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

thats why i keep getting this square root of 20 =.=

9. You need to use:

• The Distance Formula

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

which is a direct application of Pythagoras' Theorem.

• The Midpoint Formula

$(x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

10. Originally Posted by behindDkeys
oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

thats why i keep getting this square root of 20 =.=
$\sqrt{20} = 2\sqrt{5}$ is correct.

11. Originally Posted by behindDkeys
oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

thats why i keep getting this square root of 20 =.=

Doesn't matter what order you go in as long as you are consistent.

12. Originally Posted by Prove It
$\sqrt{20} = 2\sqrt{5}$ is correct.
uhm ok so whats next to that

13. Originally Posted by behindDkeys
Ok so it says on the question:

Write the equation of the following circle:
with diameter AB and given A(4,3) and B(6,-1)

If you are familiar with the properties of circle, try this method.

Let P (x. y) be any point on the circle. Then APB is a right angled triangle. AP is perpendicular to PB.
If m1 and m2 are the slopes of AP and PB then m1*m2 = -1.

So $\frac{(y - y_1)}{(x - x_1)} \times \frac{(y - y_2)}{(x - x_2)} = -1$

Expand and get the equation of the circle.

14. Originally Posted by behindDkeys
uhm ok so whats next to that

Hello behindDkeys,
You have already been given the procedures but appear to need more help.
Try this.
Plot the two points and draw AB. AB is the diameter of a circle and its midpoint is the center.Draw the slope diagram for AB.Note that the rise is 4 and the run is 2.Circle diameter is 2rad5 and radius is rad5.Using similar triangles the midpoint of AB is 5,1. Now write the circle equation as shown by others.

bjh

15. Originally Posted by bjhopper
Hello behindDkeys,
You have already been given the procedures but appear to need more help.
Try this.
Plot the two points and draw AB. AB is the diameter of a circle and its midpoint is the center.Draw the slope diagram for AB.Note that the rise is 4 and the run is 2.Circle diameter is 2rad5 and radius is rad5.Using similar triangles the midpoint of AB is 5,1. Now write the circle equation as shown by others.

bjh
sorry i forgot to post what i got

here's the equation i got:
$x^2+y^2=\sqrt{5}$

since radius is divided by 2 and the diameter i got was $2\sqrt{5}$

did i get it right?

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