A circle has a center that falls on the line #y = 5/8x +6 # and passes through # ( 5 ,2 )# and #(3 ,8 )#. What is the equation of the circle?

Question

A circle has a center that falls on the line #y = 5/8x +6 # and passes through #      (    5  ,2  )# and #(3   ,8  )#.  What is the equation of the circle?

Ella Armstrong · Accepted Answer

# (x + 8)^2 + (y-1)^2 = 170 #

The "12 minutes ago" without a comment worries me. Where have they disappeared to?

We call the center #(x,y)# and equate the squared radii, brashly writing the equation:

# (x - 5)^2 + ( (5/8 x + 6) - 2)^2 = (x - 3)^2 + ( (5/8 x + 6) - 8)^2#

# (x - 5)^2 + (5/8 x + 4)^2 = (x - 3)^2 + ( 5/8 x - 2)^2#

Let's multiply by #8^2# to clear the fractions:

# 64 (x - 5)^2 + (5 x + 32)^2 = 64(x - 3)^2 + ( 5 x - 16)^2#

Expanding, we can see that the squared terms are the same on both sides, so I won't bother writing them. After that, it appears that a factor of 32 can also be cancelled:

# 32( -20 x + 50 + 10 x + 32) = 32( - 12 x +18 - 5 x + 8) #

# 7 x = -56 #

#x = -8 #

#y = 5/8 x + 6 = 1 #

Squared radius # (-8 -5)^2 + (1 - 2)^2 = 170 #

Equation:

# (x + 8)^2 + (y-1)^2 = 170 #

Check:

# (3+8)^2 + (8-1)^2 = 11^2+7^2=121+49=170 quad sqrt #

Evelyn Arboleda · Answer

# S : x^2+y^2+16x-2y-105=0#.

Suppose that the reqd. eqn. of the circle #S# is given by, # S : x^2+y^2+2gx+2fy+c=0#. The centre #C# of #S# is #C(-g,-f) in" the line : "y=5/8*x+6#. #:. -f=-5/8*g+6, or, f=5/8g-6.........................(1)#. #(5,2) in S. :. 5^2+2^2+10g+4f+c=0, or, # # 29+10g+4f+c=0......................................................(2)#. Likewise, #(3,8) in S. :. 73+6g+16f+c=0.................(3)#. #(3)-(2) rArr 44-4g+12f=0 rArr 11-g+3f=0.........(4)#. #(1) & (4) rArr 11-g+3(5/8g-6)=0#. #:. -7+7/8g=0 rArr g=8........................................(star1)#. #:. f=5/8g-6=5/8*8-6=-1............................(star2)#. With #g=8,f=-1 and (2)#, we get, # 29+80-4+c=0 rArr c=-105.............................(star3)#. #(star1), (star2) and (star3)# give us, # S : x^2+y^2+16x-2y-105=0#.

Savannah Aguilar · Answer

# x^2+y^2+16x-2y-105=0#,

This is the Third Way to determine the circle's equation. The eqn. of a circle having #(5,2) and (3,8)# as extremities of its diameter is given by, # s : (x-5)(x-3)+(y-2)(y-8)=0#. #:. s : x^2+y^2-8x-10y+31=0#. The eqn. of the chord joining #(5,2) and (3,8)# is given by, # l : |(x,y,1),(5,2,1),(3,8,1)|=0#, # i.e., -6x-2y+34=0 or, l : 3x+y-17=0#. Now, since the reqd. circle #S# passes through the points of intersection of #s and l#, we may take, # S : s+lambdal=0, or, # # S : x^2+y^2-8x-10y+31+lambda(3x+y-17)=0#, # i.e., S : x^2+y^2+(3lambda-8)x+(lambda-10)y+31-17lambda=0#. Clearly, the centre #C# of #S# is, #C(4-3/2lambda,5-lambda/2)#, which lies on #y=5/8x+6#. #:. 5-lambda/2=5/8(4-3/2lambda)+6#. #:. lambda=8#. Thus, the eqn. for the required circle is # S : x^2+y^2-8x-10y+31+8(3x+y-17)=0#, # i.e., S : x^2+y^2+16x-2y-105=0#, as before!

Evelyn Arboleda · Answer

undefined The equation of the circle is:

(x - 7)^2 + (y - 4)^2 = 10.