Chapter 1.4, Problem 10AYU

To determine

To write: The standard form of the equation of the circle with end points of a diameter $\left(0,1\right)$ and $\left(2,3\right)$

Answer to Problem 10AYU

$\text{Centre = }\left(\text{1, 2}\right)$

$\text{Radius = }\sqrt{2}$

$\text{The standard form of the equation of the circle is }\left(x- 1\right)²+\left(y -2\right)²=2$

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 $\text{A}\left(\text{0, 1}\right)\text{ and B}\left(\text{2, 3}\right)$.

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 $\text{A(}x₁, y₁)=A\left(0,1\right)$ and $\text{B}(x₂,y₂)=B\left(2,3\right)$.

$\text{Midpoint O }=\left(x,y\right)=(\frac{0 + 2}{2}, \frac{1 + 3}{2})$

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

$=\left(1,2\right)$

Therefore, the centre of the circle is $\text{O}\left(1,2\right).$

By using distance formula:

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

$=\sqrt{{(1 - 0)}^2 + {(2 - 1)}^2}$

$=\sqrt{{(1)}^2 + {(1)}^2}$

$=\sqrt{1 + 1}$

$=\sqrt{2}$

Now substitute the values $r = \sqrt{2},h = 1,k= 2$ in the standard form of an equation of a circle.

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

$\left(x - 1\right)²+\left(y - 2\right)²=(\sqrt{2})²$

$\left(x - 1\right)²+\left(y - 2\right)²=2$