Question

What is the standard form of the equation of a circle passing through (0, -14), (-12, -14), and (0,0)?

Answer 1

A circle of radius `sqrt(85)` and centre `(-6,-7)`

Standard form equation is: `(x+6)^2 + (y+7)^2 = 85`

Or, `x^2+12x + y^2+14y = 0`

Explanation:

The Cartesian equation of a circle with centre `(a,b)` and radius `r` is:

`(x-a)^2 + (y-b)^2 = r^2`

If the circle passes through (0,-14) then:

`(0-a)^2 + (-14-b)^2 = r^2`
`a^2 + (14+b)^2 = r^2` ................................ [1]

If the circle passes through (0,-14) then:

`(-12-a)^2 + (-14-b)^2 = r^2`
`(12+a)^2 + (14+b)^2 = r^2` ................................ [2]

If the circle passes through (0,0) then:

`(0-a)^2 + (0-b)^2 = r^2`
`a^2 + b^2 = r^2` ................................ [3]

We now have 3 equations in 3 unknowns

Eq [2] - Eq [1] gives:

`(12+a)^2 -a^2 = 0`
`:. (12+a-a)(12+a+a)=0`
`:. 12(12+2a)=0`
`:. a=-6`

Subs `a=6` into Eq [3]:

`36+b^2 = r^2` ................................ [4]

Subs `a=6` and `r^2=36+b^2`into Eq [1]:

`36 + (14+b)^2 = 36+b^2`
`:. (14+b)^2 - b^2 = 0`
`:. (14+b-b)(14+b+b) = 0`
`:. 14(14+2b) = 0`
`:. b=-7`

And finally, Subs `b=-7` into Eq [4];

`36+49 = r^2`
`:. r^2 = 85`
`:. r = sqrt(85)`

And so the equation of the circle is

`(x+6)^2 + (y+7)^2 = 85`

Which represents a circle of radius `sqrt(85)` and centre `(-6,-7)`

We can multiply out if required to get:

`x^2+12x+36 + y^2+14y+49 = 85`
`x^2+12x + y^2+14y = 0`