What is the standard form of the equation of a circle passing through (0, -14), (-12, -14), and (0,0)?
1 Answer
A circle of radius
Standard form equation is:
Or,
Explanation:
The Cartesian equation of a circle with centre
(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
36+b^2 = r^2 ................................ [4]
Subs
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
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
We can multiply out if required to get:
x^2+12x+36 + y^2+14y+49 = 85
x^2+12x + y^2+14y = 0