A circle has a center that falls on the line y = 3/7x +3 and passes through ( 2 ,8 ) and (3 ,5 ). What is the equation of the circle?

1 Answer
Somebody N.
Jun 17, 2018

color(blue)((x-28)^2+(y-15)^2=725

Explanation:

The standard form of the equation of a circle is given as:

(x-h)^2+(y-k)^2=r^2

Where:

bbh and bbk are the bbx and bby co-ordintates of the centre respectively, and bbr is the radius.

If the centre lie on the line y=3/7x+3, then bbh and bbk lie on this line:

:.

k=3/7h+3 \ \ \ [1]

We are given two points on the circumference of the circle:

(2,8), (3,5)

Using: (x-h)^2+(y-k)^2=r^2

(2-h)^2+(8-k)^2=r^2 \ \ \ [2]

(3-h)^2+(5-k)^2=r^2 \ \ \ [3]

Solving simultaneously:

[2] - [3]

-5+2h+39-6k=0

34+2h-6k=0

k=1/3h+17/3 \ \ \ [4]

"substituting " [1] " in " [4]

3/7h+3=1/3h+17/3

3/7h+3-1/3h-17/3=>h=28

In [1]

k=3/7(28)+3=15

Centre of circle is at (28,15)

Using this and point (3,5), we have:

r^2=(3-28)^2+(5-15)^2=725

Equation of circle is:

color(blue)((x-28)^2+(y-15)^2=725

PLOT:

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