College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding - Page 186: 11


center: $(2.5, 2)$; radius = $1.5$ standard form of the equation: $\color{blue}{(x-2.5)^2+(y-2)^2=2.25}$

Work Step by Step

RECALL: The standard form of a circle's equation is: $(x-h)^2 +(y-k)^2=r^2$ where $r$ = radius and $(h, k)$ is the center. The points on the circle $(1, 2)$ and $(4, 2)$ belong to the same horizontal line. This means that the segment connecting them is a diameter of the circle. The center of the circle is the midpoint of the diameter. Find the midpoint using the midpoint formula $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$: center = $\left(\dfrac{1+4}{2}, \dfrac{2+2}{2}\right)=(2.5, 2)$ Substitute $2.5$ to $h$ and $2$ to $k$ in the standard form above to obtain the tentative equation: $(x-2.5)^2+(y-2)^2=r^2$ To find the value of $r$, substitute the x and y values of the point $(1, 2)$ into the tentative equation above to obtain: $(x-2.5)^2+(y-2)^2=r^2 \\(1-2.5)^2+(2-2)^2=r^2 \\(-1.5)^2+0^2=r^2 \\2.25 + 0 = r^2 \\2.25 = r^2 \\\pm\sqrt{2,25} = \sqrt{r^2} \\\pm 1.5 = r$ Since the radius cannot be negative, $r=1.5$ Therefore, the standard form of the circle's equation is: $(x-2.5)^2+(y-2)^2=(1.5)^2 \\\color{blue}{(x-2.5)^2+(y-2)^2=2.25}$
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