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)
please need help really fast
Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius ($\displaystyle r$).
Find the midpoint of the segment between these two points. This is your centre $\displaystyle (h, k)$.
The formula of the circle is
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$.
You need to use:
- The Distance Formula
$\displaystyle D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
which is a direct application of Pythagoras' Theorem.
- The Midpoint Formula
$\displaystyle (x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
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 $\displaystyle \frac{(y - y_1)}{(x - x_1)} \times \frac{(y - y_2)}{(x - x_2)} = -1$
Expand and get the equation of the circle.
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.
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