Chapter 1.4, Problem 9AYU

To determine

To write: The standard form of the equation of the circle with end points of a diameter $\left(1,2\right)$ and $\left(4,2\right)$

Answer to Problem 9AYU

$\text{Centre = }\left(\text{2}\text{.5, 2}\right)$

$\text{Radius = 1}\text{.5}$

$\text{The standard form of the equation of the circle is }\left(x-2.5\right)²+\left(y -2\right)²=2.25$

Explanation of Solution

Given:

The following figure is given:

From the figure we can see the centre is the midpoint of the diameter AB whose end points are $\text{A}\left(1,2\right)$ and $\text{B}\left(4,2\right).$

Formula used:

The standard form of an equation of a circle with radius $r$ and centre $\left(h,k\right)$ is:

$\left(x - h\right)²+\left(y - k\right)²=r²$

Distance formula:

$\text{d}\left[\text{A, B}\right]\text{ =}\sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}$

$\text{Midpoint formula M }=\left(x,y\right)=(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

Calculation:

Let $\text{A(}x₁, y₁)=A\left(1,2\right)$ and $\text{B}(x₂,y₂)=\text{ B}\left(4,2\right)$.

$\text{Midpoint O }=\left(x,y\right)=(\frac{1 + 4}{2},\frac{2 + 2}{2})$

$=(\frac{5}{2}, \frac{4}{2})$

$=\left(2.5,2\right)$

Therefore, the centre of the circle is $\text{O}\left(2.5,2\right).$

By using distance formula:

$\text{d}\left[\text{O, B}\right]\text{ =}\sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}$

$=\sqrt{{(4 - 2.5)}^2 + {(2 - 2)}^2}$

$=\sqrt{{(1.5)}^2 + {(0)}^2}$

$=\sqrt{{(1.5)}^2}$

$=1.5$

Now substitute the values $r =1.5,\mathrm{h }=2.5,\mathrm{k }= 2$ in the standard form of an equation of a circle.

$\left(x - h\right)²+\left(y - k\right)²=r²$

$\left(x - 2.5\right)²+\left(y - 2\right)²=\left(1.5\right)²$

$\left(x - 2.5\right)²+\left(y - 2\right)²=2.25$