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

Answer:

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 #