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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