# Given the equation x^2+y^2-8x-6y=0 (a) Write the equation in standard form. (b) State the center, radius, and intercepts. (c) Submit the graphshow all you work

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### 1 Answer

a) You need to remember what the standard form of equation of circle is, hence:

`(x-h)^2 + (y-k)^2 = r^2`

Notice that h and k represent the coordinates of center of circle and r denotes the radius.

The standard form of equation consists of sum of two binomials raised to square, hence you need to complete two squares in the original equation such that:

`(x^2 - 8x + 16)+(y^2 - 6y + 9) - 16 - 9 = 0`

`(x - 4)^2 + (y - 3)^2 = 25 =gt (x - 4)^2 + (y - 3)^2 = 5^2`

`` **The standard form of equation of circle is `(x - 4)^2 + (y - 3)^2 = 5^2` .**

b) The standard form of equation of circle tells what the center and radius are, hence the center of circle is c=(4,3) and the radius is r=5.

The circle intercepts x axis at y = 0, hence you need to substitute 0 for y such that:

`(x - 4)^2 + (0 - 3)^2 = 5^2 =gt (x - 4)^2 = 25 - 9`

`` `(x - 4)^2 = 16 =gt (x - 4) = +-sqrt16 =gt (x - 4) = +-4`

`x - 4 = 4 =gt x = 8 `

`x - 4 = -4 =gt x = 0`

**Hence, the circle intercepts x axis at (0,0) and (8,0).**

You need to remember that a curve intercepts y axis at x = 0, hence you need to substitute 0 for x in equation of circle such that:

`(0 - 4)^2 + (y - 3)^2 = 5^2 =gt (y - 3)^2 = 25 - 16`

`(y - 3)^2 = 9 =gt (y - 3) = +- sqrt9`

`` `(y - 3) = +- 3`

`` `y - 3 = 3 =gt y = 6`

`` `y - 3 = -3 =gt y = 0`

**Hence, the circle intercepts y axis at origin (0,0) and (0,6).**