Question #158886

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. 


1
Expert's answer
2021-01-29T12:40:46-0500

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=(x1x0)2+(y1y0)2d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}



d1=((5)2)2+(04)2=(7)2+(4)2=49+16=65d_1 = \sqrt{((-5) - 2)^2 + (0 - 4)^2} = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}

d2=(22)2+(34)2=(4)2+(1)2=16+1=17d_2 = \sqrt{(-2 - 2)^2 + (3 - 4)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}

d2=(62)2+(114)2=(4)2+(15)2=16+225=241d_2 = \sqrt{(6 - 2)^2 + (-11 - 4)^2} = \sqrt{(4)^2 + (-15)^2} = \sqrt{16 + 225} = \sqrt{241}


Secondly, by definition a circle is:

r2=(x1x0)2+(y1y0)2r^2 = (x_1 - x_0)^2 + (y_1 - y_0)^2

r=(x1x0)2+(y1y0)2=d1=d2=d3r = \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:


6517241,d1d2d3\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




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