How do you find an equation of a circle passing through points (-6,-1) , (-4,3) , (2,-5)?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

13

This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Learn more about featured answers

Jan 22, 2016

Answer:

#x^2+y^2-14x-10y=131#

Explanation:

Given 3 points

#P_1(-6.-1),P_2(-4,3),P_3(2,-5)#

For equation 1
#P_1=P_2#

#(x+6)^2+(y+1)^2=(x+4)^2+(y-3)^2#

By expanding, we have
#x^2+12x+36 +y^2+2y+1 =x^2+8x+16+y^2-6y+9#

By solving further, we get...

#4x+8y=-12#

#x+2y=-3# (Equation 1)

To solve for equation 2

We let

#P_2=P_3#

#(x+4)^2+(y-3)^2= (x-2)^2+(y+5)^2#

#x^2+8x+16+y^2-6y+9= x^2-4x+4+y^2+10y+25#

Using the same method we did on equation 1, we get...

#12x-16y=4#

#3x-4y=1# (Equation 2)

We can find the center by solving the 2 equations obtained by elimination method.

#-2[x+2y=-3]#
#3x-4y=1#

#-2x+4y=6#
#3x-4y=1#

#x=7#

#3(7)-4y=1#

#21-4y=1#
#-4y=-20#

#y=5#

We now have the center of the circle

#C(7,5)#

To solve for the radius, we use the distance formula and substitute the center and either of the three points

#r^2=(-6-7)^2+(-1-5)^2#
#r^2=(-13)^2+(-6)^2#
#r^2=169+36#

#r^2=205#

We now have the components of the circle.

#(x-7)^2+(y-5)^2=205#

By expanding the equation...

#x^2-14x+49+y^2-10y+25=205#

By combining like terms and solving, we get...
#x^2+y^2-14x-10y=131#

Was this helpful? Let the contributor know!
1500
Impact of this question
2243 views around the world
You can reuse this answer
Creative Commons License