Chapter 1.4, Problem 8AYU

To determine

To write: The standard form of the equation of the circle with centre $\left(1,2\right)$ and radius 2

Answer to Problem 8AYU

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

$\text{Radius = 2}$

$\text{The standard form of the equation of the circle is }\left(x- 1\right)²+\left(y -2\right)²=4$

Explanation of Solution

Given:

The following figure is given:

From the figure we can see the centre is $\left(1,2\right)$ and the radius is the distance between $\left(1,0\right)$ and $\left(1,2\right)$.

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:

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

Calculation:

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

By using distance formula:

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

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

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

$=\sqrt{4}$

$=2$

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

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

$\left(x -1\right)²+\left(y -2\right)²=2²$

$\left(x - 1\right)²+\left(y - 2\right)²=4$