Chapter 1.4, Problem 15AYU

To determine

To write: The standard form and general form of the equation of the circle with radius 5 and centre $\left(4,-3\right)$

Answer to Problem 15AYU

$\text{The standard form of the equation of the circle is }\left(x\text{ - 4}\right)\text{² + }\left(y\text{ + 3}\right)\text{² = 25, and general form of the equation of the circle is }x\text{² + }y\text{² - 8}x\text{ + 6}y\text{ = 0}\text{.}$

Explanation of Solution

Given:

Radius $r\text{ = 5;}$ centre $\left(h,k\right)\text{ = }\left(\text{4, −3}\right)$

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 $r\text{ = 5}$ and $\left(h,k\right)=\left(4,-3\right)$ in the standard form of the equation.

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

$\left(x-4\right)²+\left(y-\left(-3\right)\right)²=5²$

$\left(x-4\right)²+\left(y+3\right)²=25$

$x²-8x+16+y²+6y+9=25$

$x²-8x+y²+6y+25-25=25-25$   Subtract 25 from both sides

$x²+y²-8x+6y=0$

General form of the equation of the circle is $x²+y²-8x+6y=0$.