A circle has a center that falls on the line y = 5/3x +1 and passes through (8 ,2 ) and (3 ,2 ). What is the equation of the circle?

1 Answer
Shwetank Mauria
Jul 28, 2017

36x^2+36y^2-396x-732y+1095=0

Explanation:

As ordinates of points (8,2) and (3,2) are same, the line joining them is parallel to x-axis and its equation is y=2 and the midpoint of points (8,2) and (3,2) being (11/2,2), center lies on x=11/2.

Further center lies on the pont of intersection of y=5/3x+1 and x=11/2 and therefore we have

y=5/3xx11/2+1=55/6+1=61/6 and center is (11/2,61/6) and as circle passes through (3,2) the radius is

sqrt((3-11/2)^2+(2-61/6)^2)=sqrt(25/4+2401/36)=sqrt(2626/36)

and hence equation of circle is

(x-11/2)^2+(y-61/6)^2=2626/36

or x^2-11x+121/4+y^2-61/3y+3721/36=2626/36

or 36x^2+36y^2-396x-732y+2184=0

or 9x^2+9y^2-99x-183y+546=0

graph{(9x^2+9y^2-99x-183y+546)((x-8)^2+(y-2)^2-0.1)((x-3)^2+(y-2)^2-0.1)(2x-11)((x-11/2)^2+(y-61/6)^2-0.1)(3y-5x-3)=0 [-16, 24, 0, 20]}

Related questions
See all questions in Equation of a Circle
Impact of this question
1536 views around the world
You can reuse this answer
Creative Commons License