Answer to Question #158886 in Geometry for Aidan

Question

Show that the points P(–5, 0), Q(–2, 3), and R(6, –11) lie on a circle with its center at C(2, 4). Conclude by explaining why your findings show this. 

Answer 1

Solution:

First of all, let us find the distance between points P and C, Q and C and R and C. Here is the formula:

$$d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$$

$d_1 = \sqrt{((-5) - 2)^2 + (0 - 4)^2} = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$

$d_2 = \sqrt{(-2 - 2)^2 + (3 - 4)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$

$d_2 = \sqrt{(6 - 2)^2 + (-11 - 4)^2} = \sqrt{(4)^2 + (-15)^2} = \sqrt{16 + 225} = \sqrt{241}$

Secondly, by definition a circle is:

$r^2 = (x_1 - x_0)^2 + (y_1 - y_0)^2$

$r = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = d_1=d_2=d_3$

Therefore, all points should be equidistant from point C

But in our case they are not:

$\sqrt{ 65} \neq \sqrt {17} \neq \sqrt {241}, d_1 \neq d_2 \neq d_3$

Therefore, these points do not lie on a circle

Answer:

Points P, Q, R do not lie on the circle with center C(2, 4) as they are not equidistant from point C