Chapter 1.5, Problem 46AYU

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

46.

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 46AYU

Solution:

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

Explanation of Solution

Given:

The following graph:

In the graph the circle touches the $x$ -axis at $\left(-3,0\right)$ and the $y$ -axis at $\left(0,3\right)$. The radius drawn from the points $\left(-3,0\right)$ and $\left(0,3\right)$ will intersect at $\left(-3,3\right)$. This gives centre of the circle as $\left(-3,3\right)$.

Formula used:

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

Calculation:

Let $\text{A(x}₁\text{, y}₁\text{) = }\left(\text{-3, 0}\right)\text{, B (x}₂\text{, y}₂\text{) = }\left(\text{-3, 3}\right)$.

$\text{d}\left(\text{A, B}\right)=\sqrt{{(-3 - \left(-3\right))}^2 + {(3 - 0)}^2}$

$=\sqrt{{(-3 + 3)}^2 + {(3)}^2}$

$\text{= }\sqrt{\text{0 + 9}}$

$=\sqrt{9}$

$=3$

Standard equation of the circle with centre $\left(-3,3\right)$ and radius 3:

$\left(x - \left(-3\right)\right)²+\left(y - 3\right)²=\left(3\right)²$

$\left(x + 3\right)²+\left(y - 3\right)²=9$

$\text{d) }\left(x + 3\right)²+\left(y - 3\right)²=9$ is the equation of the circle given in the figure.