A circle has a center that falls on the line #y = 5/3x +1 # and passes through #(5 ,2 )# and #(3 ,2 )#. What is the equation of the circle?

1 Answer
Ratnaker Mehta
Jun 12, 2016

# 3x^2+3y^2-24x-46y+125=0.#

Explanation:

Let the given points be #A(5,2), B(3,2).#

Let us observe that the y-co-ordinates of A & B are same, hence, #AB# a is a horizontal chord.

We know that the centre of a circle lies on the perpendicular bisector (p.b.) of a chord. Chord #AB# is horizontal, so, its p.b. is a vertical line thro. the midpoint of #AB#, i.e., #x=(5+3)/2.# The cetre is on #x=4...... (i)#

By what is given, centre is also on the line #y=(5/3)x+1. .......(ii)#

Solving #(i)# and #(ii)#, #y=23/3.#

So, we get the centre #(4,23/3).#

Next goal is to find radius r of the reqd. cicle. We note that

#CA=CB=r.#

We use # CB^2# = #r^2= (4-3)^2+(23/3-2)^2=1+289/9.#

Hence, the reqd. eqn.of the circle is #(x-4)^2+(y-23/3)^2=r^2=(4-3)^2+(23/3-2)^2.#
#:. (x-4)^2+(y-23/3)^2#-- #(4-3)^2 - (23/3-2)^2.#
#:.(x-4)^2- (4-3)^2+(y-23/3)^2- (23/3-2)^2=0..#
#:.(x-4+4-3)(x-4-4+3)(y-23/3+23/3-2)(y-23/3-23/3+2)=0.#
#:. (x-3)(x-5)+(y-2)(y-40/3)=0.#
#:.3(x^2-8x+15)+(y-2)(3y-40)=0.#
#:. 3x^2-24x+45+3y^2-46y+80=0.#
#:. 3x^2+3y^2-24x-46y+125=0.#

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