Chapter 1.5, Problem 47AYU

In Problems 45-48, match each graph with the correct equation.

47.

a. ${\left(x-3\right)}^2+{\left(y\text{ + 3}\right)}^2=9$

b. ${\left(x+\text{ 1}\right)}^2+{\left(y-2\right)}^2=4$

c. ${\left(x-1\right)}^2+{\left(y+2\right)}^2=4$

d. ${\left(x+3\right)}^2+{\left(y-3\right)}^2=9$

To determine

To match: The given graph with the correct equation given in options

Answer to Problem 47AYU

Solution:

$\text{b) }\left(x + 1\right)²+\left(y - 2\right)²=4$

Explanation of Solution

Given:

The following graph:

In the graph the circle touches the $x$ -axis at $\left(-1,0\right)$ and the end points of the diameter are $\left(-1,0\right)$ and $\left(-1,4\right)$. The diameter is parallel to negative $y$ -axis.

Formula used:

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

$\text{D(P}₁\text{, P}₂\text{) = }\sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}$

Calculation:

Midpoint of the diameter is the centre of the circle.

Let $\text{A}(x₁,y₁)=\left(-1,0\right),\text{ B }(x₂,y₂)=\left(-1,4\right)$.

Midpoint $\text{M }=\left(x,y\right)=(\frac{\left(-1\right) + \left(-1\right)}{2},\frac{0 + 4}{2})$

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

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

Centre of the circle is $\left(-1,2\right)$.

$\text{d}\left(\text{A, M}\right)=\sqrt{{(-1 - \left(-1\right))}^2 + {(2 - 0)}^2}$

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

$=\sqrt{0 + 4}$

$=\sqrt{4}$

$\text{= 2}$

Standard equation of the circle with centre $\left(-1,2\right)$ and radius 2:

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

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

$\text{b) }\left(x + 1\right)²+\left(y - 2\right)²=4$ is the equation of the circle given in the figure.