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

1 Answer
Jan 14, 2017

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^2into 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