Chapter 1.4, Problem 16AYU

To determine

To write: The standard form and general form of the equation of the circle with radius 4 and centre $\left(2,-3\right)$

Answer to Problem 16AYU

$\text{The standard form of the equation of the circle is }\left(x\text{ − 2}\right)\text{² + }\left(y\text{ + 3}\right)\text{² = 16, and general form of the equation of the circle is }x\text{² + }y\text{² − 4}x\text{ + 6}y\text{ − 3 = 0}\text{.}$

Explanation of Solution

Given:

Radius $r\text{ = 4;}$ centre $\left(h,k\right)\text{ = }\left(\text{2, −3}\right)$

Formula used:

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

$\left(x\text{ − }h\right)\text{² + }\left(y\text{ − }k\right)\text{² = }r\text{²}$

Calculation:

Substitute $r\text{ = 4}$ and $\left(h,k\right)=\left(2,-3\right)$ in the standard form of the equation.

$\left(x\text{ − }h\right)\text{² + }\left(y\text{ − }k\right)\text{² = }r\text{²}$

$\left(x - 2\right)²+\left(y -\left(-3\right)\right)²=4²$

$\left(x-2\right)²+\left(y+3\right)²=16$

$x²-4x+4+y²+6y+9=16$

$x²-4x+y²+6y+13-16=16-16$   Subtract 16 from both sides

$x²+y²-4x+6y\text{ − 3 }=0$

General form of the equation of the circle is $x² + y²-4x + 6y -3=0$.